# How do you solve x^2-8x=-10 using the quadratic formula?

Aug 17, 2017

$x = 4 + \sqrt{11} \mathmr{and} x = 4 - \sqrt{11}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 8 x = - 10$
Re-write in standard form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{1} {x}^{2} \textcolor{b l u e}{- 8} x + \textcolor{g r e e n}{10} = 0$

The quadratic formula tells us that an equation of the form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
has solutions:
color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b-4color(red)acolor(green)c))/(2color(red)a

So in this case the solutions are
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \left(\textcolor{b l u e}{- 8}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 8}\right)}^{2} - 4 \times \textcolor{red}{1} \times \textcolor{g r e e n}{10}}}{2 \times \textcolor{red}{1}}$

$\textcolor{w h i t e}{\text{XXXx}} = \frac{8 \pm \sqrt{44}}{2}$

$\textcolor{w h i t e}{\text{XXXx}} = \frac{8 \pm 2 \sqrt{11}}{2}$

$\textcolor{w h i t e}{\text{XXXx}} = 4 \pm \sqrt{11}$