{"id":676,"date":"2013-05-28T14:28:16","date_gmt":"2013-05-28T14:28:16","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=676"},"modified":"2026-03-25T10:39:22","modified_gmt":"2026-03-25T15:39:22","slug":"using-the-quadratic-formula","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/using-the-quadratic-formula\/","title":{"rendered":"Using the Quadratic Formula"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_C_HiqzYRW1U\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_C_HiqzYRW1U\" data-source-videoID=\"C_HiqzYRW1U\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Using the Quadratic Formula Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Using the Quadratic Formula\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_C_HiqzYRW1U:hover {cursor:pointer;} img#videoThumbnailImage_C_HiqzYRW1U {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/695-thumb-final-3.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_C_HiqzYRW1U\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_C_HiqzYRW1U\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Using the Quadratic Formula\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_C_HiqzYRW1U\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_C_HiqzYRW1U\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_C_HiqzYRW1U\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction Gbl_Function() {\n  var x = document.getElementById(\"Gbl\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"Gbl_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"Gbl\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#What_is_a_Quadratic_Equation\" class=\"smooth-scroll\">What is a Quadratic Equation?<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Using_the_Quadratic_Formula_to_Find_Solutions\" class=\"smooth-scroll\">Using the Quadratic Formula to Find Solutions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#The_Discriminant\" class=\"smooth-scroll\">The Discriminant<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#More_Quadratic_Formula_Examples\" class=\"smooth-scroll\">More Quadratic Formula Examples<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Quadratic_Formula_Practice_Questions\" class=\"smooth-scroll\">Quadratic Formula Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hello, and welcome to this video about using the quadratic formula!<\/p>\n<p>Today, we\u2019ll learn what the quadratic formula is, when you use it, and how it\u2019s applied to find roots in quadratic equations. <\/p>\n<h2><span id=\"What_is_a_Quadratic_Equation\" class=\"m-toc-anchor\"><\/span>What is a Quadratic Equation?<\/h2>\n<p>\nLet\u2019s start by talking about quadratic equations. A quadratic equation in standard form is written as \\(ax^{2}+bx+c=0\\), where \\(a\\neq 0\\) and \\(a\\),\\(b\\), and \\(c\\) are all real numbers. <\/p>\n<h3><span id=\"Quadratic_Formula\" class=\"m-toc-anchor\"><\/span>Quadratic Formula<\/h3>\n<p>\nThe quadratic formula is an algebraic equation that helps solve any quadratic equation. The formula is: <\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b+\\sqrt{b^{2}-4ac}}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nBy substituting in given values for the variables \\(a\\),\\(b\\), and \\(c\\), you can solve for \\(x\\) using the quadratic formula. Solving for \\(x\\) gives us <strong>solutions<\/strong> for the quadratic equation. These solutions are also called <strong>roots<\/strong>. <\/p>\n<h2><span id=\"Using_the_Quadratic_Formula_to_Find_Solutions\" class=\"m-toc-anchor\"><\/span>Using the Quadratic Formula to Find Solutions<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s consider the quadratic equation \\(x^{2}+x-6=0\\) and use the quadratic formula to find solutions for \\(x\\). <\/p>\n<p>First, identify the values for the variables \\(a\\), \\(b\\), and \\(c\\) in the quadratic equation. Since the first term, \\(x^{2}\\), is the same as \\(1x^{2}\\), the value for a is equal to 1. In the second term, \\(x\\), the value for \\(b\\) is also equal to the coefficient of 1. Since the last term is -6, the value for \\(c\\) is equal to -6. <\/p>\n<div class=\"examplesentence\">\\(a=1\\)<br \/>\n\\(b=1\\)<br \/>\n\\(c=-6\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, substitute the values for \\(a\\), \\(b\\), and \\(c\\) into the quadratic formula. So let\u2019s write out our quadratic formula again so that we remember it.<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b_{-}^{+}\\sqrt{b^{2}-4ac}}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now we\u2019re going to plug in our \\(a\\), \\(b\\), and \\(c\\) into this equation.<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( 1 \\right )_{-}^{+}\\sqrt{\\left ( 1 \\right )^{2}-4\\left ( 1 \\right )\\left ( -6 \\right )}}{2\\left ( 1 \\right )}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom here, solve for \\(x\\) using the quadratic formula. So all we have to do is simplify. So we\u2019re going to leave the -1 here for now and move on to the next part. So we have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( 1 \\right )_{-}^{+}\\sqrt{ 1-\\left ( -24 \\right )}}{2\\left ( 1 \\right )}\\)<\/div>\n<p>\n&nbsp;<br \/>\n1-(-24) is the same as saying 1+24, which is 25.<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{- 1 _{-}^{+}\\sqrt{25}}{2\\left ( 1 \\right )}\\)<br \/>\n\\(x=\\frac{-1_{-}^{+}5}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now we\u2019re going to break this apart into two different equations. So we\u2019re going to have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-1 +5}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, \\(x\\) is equal to this, so this is one of our values for \\(x\\).<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{4}{2}=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then for our other value of \\(x\\), we\u2019re going to do:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-1-5}{2}=\\frac{-6}{2}=-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the roots for this quadratic equation are 2 and -3.<\/p>\n<h2><span id=\"The_Discriminant\" class=\"m-toc-anchor\"><\/span>The Discriminant<\/h2>\n<p>\nIn the example we just looked at, there were two possible solutions, or roots, for \\(x\\). Quadratic functions can have two solutions, one solution, or no real solutions for \\(x\\).<\/p>\n<p>To figure out how many solutions a quadratic equation has, look at the discriminant in the quadratic formula. The discriminant is the portion of the quadratic formula that\u2019s beneath the square root symbol. It\u2019s the part that\u2019s read as \\(b^{2}-4ac\\). <\/p>\n<p>When simplified, the discriminant can be a positive number, a negative number, or zero, and its sign tells us how many solutions the quadratic equation has.<\/p>\n<ul>\n<li>When the discriminant is positive, then the quadratic equation has two real solutions.<\/li>\n<li>When the discriminant is negative, then the quadratic equation has no real solutions.<\/li>\n<li>When the discriminant is zero, then the quadratic equation has exactly one real solution.<\/li>\n<\/ul>\n<p>In the previous example, the discriminant was 25. Since it\u2019s a positive number, we know that the quadratic equation has two real solutions, which is what we found when we worked it out.<\/p>\n<h2><span id=\"More_Quadratic_Formula_Examples\" class=\"m-toc-anchor\"><\/span>More Quadratic Formula Examples<\/h2>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nConsider the quadratic equation \\(10x^{2}-9x+6=0\\). And let\u2019s use the quadratic formula to find solutions for \\(x\\). <\/p>\n<p>First, we need to identify the values for the variables \\(a\\), \\(b\\), and \\(c\\) in the quadratic equation. Our first term, \\(10x^{2}\\), tells us that a is equal to 10. Our second term, \\(-9x\\), tells us that the value of \\(b\\) is \\(-9\\). And our third term, \\(+6\\), tells us that \\(c\\) is equal to 6. <\/p>\n<div class=\"examplesentence\">\\(a=10\\)<br \/>\n\\(b=-9\\)<br \/>\n\\(c=6\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we\u2019re going to use the quadratic equation to solve this problem. So our formula is:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b_{-}^{+}\\sqrt{b^{2}-4ac}}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow remember, writing out formulas helps you remember it, so whenever you can, it&#8217;s a really helpful thing to do. Now all we have to do is plug in our numbers for our variables.<\/p>\n<p>So, \\(-b\\), \\(b\\) is \\(-9\\), so be careful when plugging it in. You need to make sure you have the negative \\(-9\\).<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( -9 \\right )+\\sqrt{\\left ( -9 \\right )^{2}-4\\left ( 10 \\right )\\left ( 6 \\right )}}{2\\left ( 10 \\right )}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe are going to have a positive 9 right here.<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{9_{-}^{+}\\sqrt{81-240}}{20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this gives us:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{9_{-}^{+}\\sqrt{-159}}{20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd we can stop right here because we see our discriminant is -159. Negative numbers don\u2019t have real square roots because the square is either positive or 0.<\/p>\n<p>So remember, if the discriminant is a negative number, then there are no real solutions to the equation, so we can stop right here. Our answer is that there are no real solutions for \\(x\\).<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nNow it\u2019s your turn. I\u2019m going to give you a quadratic equation, and you need to use the quadratic formula to identify possible roots for \\(x\\)<\/p>\n<div class=\"examplesentence\">\\(9x^{2}+12x+4=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nPause the video here and try it yourself. And when you\u2019re done, resume the video and we\u2019ll go over everything together.<\/p>\n<p>Let\u2019s take a look at this one together. Start by identifying the values for \\(a\\), \\(b\\), and \\(c\\). Based on the equation, we see that \\(a=9\\), \\(b=12\\), and \\(c=4\\). <\/p>\n<p>Next, substitute the values for \\(a\\), \\(b\\), and \\(c\\) into the quadratic formula: <\/p>\n<p>So remember, the formula is:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b_{-}^{+}\\sqrt{b^{2}-4ac}}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we just take our numbers up here and plug them in for all our variables. So we have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( 12 \\right )_{-}^{+}\\sqrt{\\left ( 12 \\right )^{2}-4\\left ( 9 \\right )\\left ( 4 \\right )}}{2\\left ( 9 \\right )}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this simplifies to:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-12_{-}^{+}\\sqrt{\\left ( 12 \\right )^{2}-4\\left ( 9 \\right )\\left ( 4 \\right )}}{18}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo then, we have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-12_{-}^{+}\\sqrt{0}}{18}\\)<\/div>\n<p>\n&nbsp;<br \/>\n\\(\\sqrt{0}=0\\), so all we have is:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-12_{-}^{+}0}{18}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd if you add or subtract 0, that\u2019s not going to change your number at all, so this is really the same as:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-12}{18}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich we can then simplify by dividing both the numerator and denominator by 3, so that&#8217;ll give us:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-12\\div 3}{18\\div 3}=\\frac{-4}{6}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd we can simplify that one more time by dividing both by 2 to get:<\/p>\n<div class=\"examplesentence\">\\(x=-\\frac{2}{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(x=-\\frac{2}{3}\\). And that\u2019s our root for this equation.<\/p>\n<h3><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nQuadratic functions occur in many real-world situations. One example is shooting a cannonball out of a cannon. When you do this, it goes up and then it comes back down in the shape of a parabola. Since a parabola is the graph of a quadratic function, you can use the equation for the path the cannonball follows to answer questions about its time, speed, and distance.<\/p>\n<p>There are lots of other problem-solving instances in which quadratic equations are helpful. Understanding them helps us apply mathematical concepts in the real world. <\/p>\n<p>I have one more problem for you to try. Let\u2019s take a look at a story problem that we can solve by using the quadratic formula: <\/p>\n<div class=\"transcriptcallout\" style=\"text-align: left;\">A high school physics club shoots a model rocket into the air from 12 feet above the ground. The initial velocity of the rocket is 19 feet per second. The equation \\(h=-10x^{2}+19x+12\\) can be used to show the height of the rocket after \\(x\\) seconds. How long does it take for the rocket to hit the ground?<\/div>\n<p>\n&nbsp;<br \/>\nWe can solve this problem by applying the quadratic formula to the equation given in the scenario. This problem is a little bit more challenging, but I know you can handle it! Pause the video here and try it yourself. And when you\u2019re done, we\u2019ll go over it together. <\/p>\n<p>Let\u2019s take a look at this one.<\/p>\n<p>In this scenario, we are looking for the amount of time it takes for the rocket to reach the ground. The height (\\(h\\)) at ground level is 0 feet. Therefore, we can replace 0 for \\(h\\) in the equation. So we can rewrite it as:<\/p>\n<div class=\"examplesentence\">\\(0=-10x^{2}+19x+12\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now it looks like our quadratic equation in standard form, just with our 0 on the left side instead of the right side. So now that we understand that our equation is really just in standard form, we can identify the values for \\(a\\), \\(b\\), and \\(c\\). Based on our equation, we see that \\(a=-10\\), \\(b=19\\), and \\(c=12\\). <\/p>\n<p>Now we can substitute these values into our quadratic formula. Here is the formula: <\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-b_{-}^{+}\\sqrt{b^{2}-4ac}}{2a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNo we can plug in our numbers. So we have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( 19 \\right )^{2}-4\\left ( -10 \\right )\\left ( 12 \\right )}{2\\left ( -10 \\right )}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this is equal to:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 1.5px 1.5px 5px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(x=\\frac{-19_{-}^{+}\\sqrt{361-\\left ( -480 \\right )}}{-20}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So this is equal to:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-\\left ( 19 \\right )+\\sqrt{841}}{-20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this is equal to:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-19_{-}^{+}29}{-20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now from this step, we&#8217;re going to take this equation and break it into two separate equations because of this plus or minus. So we\u2019re going to have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-19-29}{-20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd we\u2019re also going to have:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-19-29}{-20}\\)<br \/>\n\\(x=\\frac{-19+29}{-20}=\\frac{10}{-20}=-\\frac{1}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd over here:<\/p>\n<div class=\"examplesentence\">\\(x=\\frac{-19-29}{-20}=\\frac{-48}{-20}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur negatives will cancel each other out, and if we divide both the numerator and denominator by 4, we&#8217;ll get:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 1.5px 1.5px 5px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(x=\\frac{-48}{-20}=\\frac{48\\div 4}{20\\div 4}=\\frac{12}{5}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So the solutions to this equation are \\(x=-\\frac{1}{2}\\) and \\(x=\\frac{12}{5}\\).<\/p>\n<p>Now that we have the two roots for \\(x\\), we must interpret what they mean in the context of the word problem.<\/p>\n<p>Recall that \\(x\\) stands for seconds, measuring the amount of time it takes for the model rocket to hit the ground. Time is measured with positive numbers, not negative numbers. Therefore, the negative root of \\(x\\) does not apply to this scenario. Based on the positive solution for \\(x\\), the amount of time it will take for the rocket to hit the ground is \\(\\frac{12}{5}\\) seconds. <\/p>\n<p>Great work! <\/p>\n<p>I hope this video about using the quadratic formula was helpful. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Quadratic_Formula_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Quadratic Formula Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the equation using the quadratic formula:<\/p>\n<div class=\"yellow-math-quote\">\\(2x^2-5x-3=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x=\\{-\\frac{1}{4},4\\}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">\\(x=\\{3,-\\frac{1}{2}\\}\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(x=\\{-6,3\\}\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(x=\\{2,-3\\}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Plug \\(a\\), \\(b\\), and \\(c\\) into the quadratic formula:<\/p>\n<p style=\"text-align: center\">\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\)<\/span><\/p>\n<ul>\n<li>\\(a\\) represents the coefficient in front of \\(x^2\\) (in this case, \\(a=2\\))<\/li>\n<li>\\(b\\) represents the coefficient in front of \\(x\\) (in this case, \\(b=-5\\))<\/li>\n<li>\\(c\\) represents the constant (in this case, \\(c=-3\\))<\/li>\n<\/ul>\n<p style=\"text-align:center; line-height: 50px;\">\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\:\\frac{-(-5)\\pm\\sqrt{(-5)^2-4(2)(-3)}}{2(2)}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{5\\pm\\sqrt{25+24}}{4}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{5\\pm\\sqrt{49}}{4}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{5\\pm7}{4}\\)<\/span><br \/>\n\\(x=\\frac{5+7}{4}\\:\\:\\:\\:\\:\\:x=\\frac{5-7}{4}\\)<br \/>\n\\(x=3\\:\\:\\:\\:\\:\\:\\:\\:\\:x=\\frac{-1}{2}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the equation using the quadratic formula:<\/p>\n<div class=\"yellow-math-quote\">\\(2x^2+7x-4=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(x=\\{\\frac{1}{2},5\\}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(x=\\{\\frac{1}{4},4\\}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(x=\\{4,-5\\}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\(x=\\{\\frac{1}{2},-4\\}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Plug \\(a\\), \\(b\\), and \\(c\\) into the quadratic formula:<\/p>\n<p style=\"text-align: center\">\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\)<\/span><\/p>\n<ul>\n<li>\\(a\\) represents the coefficient in front of \\(x^2\\) (in this case, \\(a=2\\))<\/li>\n<li>\\(b\\) represents the coefficient in front of \\(x\\) (in this case, \\(b=7\\))<\/li>\n<li>\\(c\\) represents the constant (in this case, \\(c=-4\\))<\/li>\n<\/ul>\n<p style=\"text-align:center; line-height: 50px;\">\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-7\\pm\\sqrt{7^2-4(2)(-4)}}{2(2)}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-7\\pm\\sqrt{49+32}}{4}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-7\\pm\\sqrt{81}}{4}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-7\\pm9}{4}\\)<\/span><br \/>\n\\(x=\\frac{-7+9}{4}\\:\\:\\:\\:\\:\\:x=\\frac{-7-9}{4}\\)<br \/>\n\\(x=\\frac{1}{2}\\:\\:\\:\\:\\:\\:\\:\\:\\:x=-4\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the equation using the quadratic formula:<\/p>\n<div class=\"yellow-math-quote\">\\(9x^2-6x+1=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x=4\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(x=\\frac{1}{2}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-3\">\\(x=\\frac{1}{3}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(x=7\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Plug \\(a\\), \\(b\\), and \\(c\\) into the quadratic formula:<\/p>\n<p style=\"text-align: center\">\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\)<\/span><\/p>\n<ul>\n<li>\\(a\\) represents the coefficient in front of \\(x^2\\) (in this case, \\(a=9\\))<\/li>\n<li>\\(b\\) represents the coefficient in front of \\(x\\) (in this case, \\(b=-6\\))<\/li>\n<li>\\(c\\) represents the constant (in this case, \\(c=1\\))<\/li>\n<\/ul>\n<p style=\"text-align:center; line-height: 55px;\">\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-(-6)\\pm\\sqrt{(-6)^2-4(9)(1)}}{2(9)}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{6\\pm\\sqrt{36-36}}{18}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{6\\pm\\sqrt{0}}{18}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{6}{18}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{1}{3}\\)<\/span><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the equation using the quadratic formula:<\/p>\n<div class=\"yellow-math-quote\">\\(2x^2-7x+12=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(x=\\{-2,3\\}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(x=\\{2,3\\}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">\\(x=\\:\\) no real solutions<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(x=\\{-47,7\\}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><ul>\n<li>\\(a\\) represents the coefficient in front of \\(x^2\\) (in this case, \\(a=2\\))<\/li>\n<li>\\(b\\) represents the coefficient in front of \\(x\\) (in this case, \\(b=-7\\))<\/li>\n<li>\\(c\\) represents the constant (in this case, \\(c=12\\))<\/li>\n<\/ul>\n<p style=\"text-align: center; line-height: 58px;\">\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-(-7)\\pm\\sqrt{(-7)^2-4(2)(12)}}{2(2)}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{7\\pm\\sqrt{49-96}}{4}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{7\\pm\\sqrt{-47}}{4}\\)<\/span><\/p>\n<p style=\"text-align: center; margin-top: -1em\">\\(x=\\:\\) no real solutions<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the equation using the quadratic formula:<\/p>\n<div class=\"yellow-math-quote\">\\(x^2-16x+64=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">\\(x=8\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(x=7\\)<\/div><div class=\"PQ\"  id=\"PQ-5-3\">\\(x=9\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(x=64\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><ul>\n<li>\\(a\\) represents the coefficient in front of \\(x^2\\) (in this case, \\(a=1\\))<\/li>\n<li>\\(b\\) represents the coefficient in front of \\(x\\) (in this case, \\(b=-16\\))<\/li>\n<li>\\(c\\) represents the constant (in this case, \\(c=64\\))<\/li>\n<\/ul>\n<p style=\"text-align:center; line-height: 58px;\">\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{-(-16)\\pm\\sqrt{(-16)^2-4(1)(64)}}{2(1)}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{16\\pm\\sqrt{256-256}}{2}\\)<\/span><br \/>\n\\(x=\\)<span style=\"font-size: 120%\">\\(\\:\\frac{16\\pm\\sqrt{0}}{2}\\)<\/span><br \/>\n\\(x=8\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":98461,"parent":0,"menu_order":30,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-676","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-math-advertising-group","7":"page_category-quadratics-videos","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/676","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=676"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/676\/revisions"}],"predecessor-version":[{"id":278917,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/676\/revisions\/278917"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/98461"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=676"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}