Given a quadratic equation of the following form,

$a{x}^{2}+bx+c=0$

the roots can be found using the formula below.

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Using the above formula is probably the easiest and most straightforward way to solve or find the roots of a quadratic equation.

Examples:

1. Find the roots of the quadratic equation

$0={x}^{2}-4x+3$

We subsitute the values of the coefficients, $a$, $b$ and $c$ to the formula to find $x$.

In this case, $a=1$, $b=-4$ and $c=3$. So

$\begin{array}{rl}x& =\frac{-\left(-4\right)±\sqrt{{\left(-4\right)}^{2}-4\cdot 1\cdot 3}}{2\cdot 1}\\ x& =\frac{4±\sqrt{16-12}}{2}\\ x& =\frac{4±\sqrt{4}}{2}\\ x& =\frac{4±2}{2}\\ {x}_{1}& =\frac{4-2}{2}=1\\ {x}_{2}& =\frac{4+2}{2}=3\end{array}$
2. Find the roots of the quadratic equation

$0={x}^{2}-6x+9$

Similar to the first example, we subsitute the values of the coefficients, $a$, $b$ and $c$ to the formula to find $x$.

In this case, $a=1$, $b=-6$ and $c=9$. So

$\begin{array}{rl}x& =\frac{-\left(-6\right)±\sqrt{{\left(-6\right)}^{2}-4\cdot 1\cdot 9}}{2\cdot 1}\\ x& =\frac{6±\sqrt{36-36}}{2}\\ x& =\frac{6±\sqrt{0}}{2}\\ x& =\frac{6±0}{2}\\ {x}_{1\text{,}2}& =\frac{6}{2}\\ {x}_{1\text{,}2}& =3\end{array}$

This quadratic equation has only 1 root as expected because the discriminant ( ${b}^{2}-4ac$ ) is $0$.

3. Find the roots of the quadratic equation

$0=2{x}^{2}+2x+5$

Similar to the above examples, we subsitute the values of the coefficients, $a$, $b$ and $c$ to the formula to find $x$.

In this case, $a=2$, $b=2$ and $c=5$. So

$\begin{array}{rl}x& =\frac{-2±\sqrt{{2}^{2}-4\cdot 2\cdot 5}}{2\cdot 2}\\ x& =\frac{-2±\sqrt{4-40}}{4}\\ x& =\frac{-2±\sqrt{-36}}{4}\\ x& =\frac{-2±6\sqrt{-1}}{4}\\ x& =\frac{-2±6i}{4}\\ x& =\frac{-1±3i}{2}\\ {x}_{1}& =\frac{-1-3i}{2}\\ {x}_{2}& =\frac{-1+3i}{2}\end{array}$

This quadratic equation has 2 complex roots as expected because the discriminant ( ${b}^{2}-4ac$ ) is negative.

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