# How do you solve x^2 + 8x=0 using the quadratic formula?

Aug 28, 2017

See a solution process below:

#### Explanation:

We can rewrite the equation in standard form as:

${x}^{2} + 8 x + 0 = 0$

Now, we can 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 $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{0}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{8} \pm \sqrt{{\textcolor{b l u e}{8}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{0}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- \textcolor{b l u e}{8} \pm \sqrt{64 - 0}}{2}$

$x = \frac{- \textcolor{b l u e}{8} \pm \sqrt{64}}{2}$

$x = \frac{- \textcolor{b l u e}{8} - 8}{2}$ and $x = \frac{- \textcolor{b l u e}{8} + 8}{2}$

$x = - \frac{16}{2}$ and $x = \frac{0}{2}$

$x = - 8$ and $x = 0$

=========================================

A simple way to solve this equation without the quadratic is to factor an $x$ out of each term on the left side of the equation:

$x \left(x + 8\right) = 0$

Then solve each term on the left for $0$:

Solution 1:

$x = 0$

Solution 2:

$x + 8 = 0$

$x + 8 - \textcolor{red}{8} = 0 - \textcolor{red}{8}$

$x + 0 = - 8$

$x = - 8$