# How do you solve 8x=x^2-9 using the quadratic formula?

Mar 10, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{8 x}$ from each side of the equation to put the equation in standard quadratic form:

$8 x - \textcolor{red}{8 x} = {x}^{2} - \textcolor{red}{8 x} - 9$

$0 = {x}^{2} - 8 x - 9$

${x}^{2} - 8 x - 9 = 0$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 8}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 9}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 8} \pm \sqrt{{\textcolor{b l u e}{- 8}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 9}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{8 \pm \sqrt{64 - \left(- 36\right)}}{2}$

$x = \frac{8 \pm \sqrt{64 + 36}}{2}$

$x = \frac{8 - \sqrt{100}}{2}$; $x = \frac{8 + \sqrt{100}}{2}$

$x = \frac{8 - 10}{2}$; $x = \frac{8 + 10}{2}$

$x = \frac{- 2}{2}$; $x = \frac{18}{2}$

$x = - 1$; $x = 9$

The Solution Set Is:

$x = \left\{- 1 , 9\right\}$