# How do you use the quadratic formula to solve for x-intercepts x^2 - 8x + 12 = 0?

Aug 7, 2017

See a solution process below:

#### Explanation:

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

$\textcolor{g r e e n}{12}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 8\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 8\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{12}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{\textcolor{b l u e}{8} \pm \sqrt{\textcolor{b l u e}{64} - 48}}{2}$

$x = \frac{\textcolor{b l u e}{8} \pm \sqrt{16}}{2}$

$x = \frac{\textcolor{b l u e}{8} - 4}{2}$ and $x = \frac{\textcolor{b l u e}{8} + 4}{2}$

$x = \frac{4}{2}$ and $x = \frac{12}{2}$

$x = 2$ and $x = 6$