# How do you solve 2(6 - x) = x(x + 5)  using the quadratic formula?

Jun 6, 2016

$x = \frac{- 7 \pm \sqrt{97}}{2}$

#### Explanation:

First let's operate to have the standard form:
$12 - 2 x = {x}^{2} + 5 x$
${x}^{2} + 7 x - 12 = 0$

$x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot \left(- 12\right)}}{2}$
$x = \frac{- 7 \pm \sqrt{49 + 48}}{2}$
$x = \frac{- 7 \pm \sqrt{97}}{2}$

Jun 6, 2016

$x ' = \frac{- 7 + \sqrt{97}}{2}$

$x ' = \frac{- 7 - \sqrt{97}}{2}$

#### Explanation:

$2 \left(6 - x\right) = x \left(x + 5\right)$

$12 - 2 x = {x}^{2} + 5 x$

${x}^{2} + 5 x + 2 x - 12 = 0$

${x}^{2} + 7 x - 12 = 0$

$a {x}^{2} + b x + c = 0$

The discriminant is determined using the following formula:

$\textcolor{red}{\Delta = {b}^{2} - 4 a c}$

a=1 \ ; \ b= 7 ; \ c=12

$\Delta = {\left(7\right)}^{2} - 4 \times 1 \times \left(- 12\right)$

$\Delta = 97$

Since $\Delta > 0 \implies$ two real roots exist

$\textcolor{red}{x ' = \frac{- b + \sqrt{\Delta}}{2 a}} \text{ and } \textcolor{red}{x ' ' = \frac{- b - \sqrt{\Delta}}{2 a}}$

$x ' = \frac{- 17 + \sqrt{97}}{2} \cong 1.42$

$x ' ' = \frac{- 17 - \sqrt{97}}{2} \cong - 8.42$