Use the quadratic formula calculator below to solve quadratic equations and find the value or values of \(x\).
The quadratic formula can be used to solve any quadratic equation written in standard form. The examples below show how to identify \(a\), \(b\), and \(c\), substitute them into the formula, and simplify the result.
Quadratic Formula Example 1
The quadratic formula is \(x=\large{\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\).
First, identify the values of \(a\), \(b\), and \(c\) from the equation.
\(a=1,\: b=5,\: c=6\)
Substitute these values into the quadratic formula.
\(x=\large{\frac{-5\pm\sqrt{5^2-4(1)(6)}}{2(1)}}\)
Simplify the expression under the square root.
\(x=\large{\frac{-5\pm\sqrt{25-24}}{2}}\)
\(x=\large{\frac{-5\pm1}{2}}\)
Now solve using both the plus and minus signs.
\(x=\large{\frac{-5+1}{2}}\normalsize{\:=-2}\)
\(x=\large{\frac{-5-1}{2}}\normalsize{\:=-3}\)
Therefore, the solutions are \(x=-2\) and \(x=-3\).
Quadratic Formula Example 2
For this equation, \(a=1,\: b=-6,\: c=9\).
Substitute these values into the quadratic formula.
\(x=\large{\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(9)}}{2(1)}}\)
Simplify.
\(x=\large{\frac{6\pm\sqrt{36-36}}{2}}\)
\(x=\large{\frac{6\pm0}{2}}\)
\(x=3\)
Because the expression under the square root equals zero, the equation has one repeated real solution.
Therefore, \(x=3\).
Quadratic Formula Example 3
For this equation, \(a=1,\: b=1,\: c=-1\).
Substitute these values into the quadratic formula.
\(x=\large{\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2(1)}}\)
Simplify the expression under the square root.
\(x=\large{\frac{-1\pm\sqrt{1+4}}{2}}\)
\(x=\large{\frac{-1\pm\sqrt{5}}{2}}\)
Since 5 has no perfect-square factors other than 1, \(\sqrt{5}\) cannot be simplified further, so the exact solutions remain in radical form.
Therefore, the solutions are \(x=\large{\frac{-1+\sqrt{5}}{2}}\) and \(x=\large{\frac{-1-\sqrt{5}}{2}}\).
