# How do you solve 4x^2-64=0 using the quadratic formula?

Aug 19, 2017

See a solution process below:

#### Explanation:

We can rewrite the equation as:

$4 {x}^{2} + 0 x - 64 = 0$

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:

$\textcolor{red}{4}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{0}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 64}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{0} \pm \sqrt{{\textcolor{b l u e}{0}}^{2} - \left(4 \cdot \textcolor{red}{4} \cdot \textcolor{g r e e n}{- 64}\right)}}{2 \cdot \textcolor{red}{4}}$

$x = \pm \frac{\sqrt{0 - \left(- 1024\right)}}{8}$

$x = \pm \frac{\sqrt{1024}}{8}$

$x = \pm \frac{32}{8}$

$x = \pm 4$