# How do you solve  2x^2 - 12x + 7 = 5?

Mar 28, 2018

$x = 3 \pm 2 \sqrt{2}$

#### Explanation:

Given:

$2 {x}^{2} - 12 x + 7 = 5$

Subtract $5$ from both sides to get:

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

Divide both sides by $2$ to get:

$0 = {x}^{2} - 6 x + 1$

$\textcolor{w h i t e}{0} = {x}^{2} - 6 x + 9 - 8$

$\textcolor{w h i t e}{0} = {\left(x - 3\right)}^{2} - {\left(2 \sqrt{2}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 3\right) - 2 \sqrt{2}\right) \left(\left(x - 3\right) + 2 \sqrt{2}\right)$

$\textcolor{w h i t e}{0} = \left(x - 3 - 2 \sqrt{2}\right) \left(x - 3 + 2 \sqrt{2}\right)$

Hence:

$x = 3 \pm 2 \sqrt{2}$

Mar 29, 2018

$x = 3 \pm 2 \sqrt{2}$

#### Explanation:

2x^2 - 12x + 2 = 0
x^2 - 6x + 1 = 0
Use the improved quadratic formula (Socratic search):
$D = {d}^{2} = {b}^{2} - 4 a c = 36 - 4 = 32 = 2 \left(16\right)$
$d = \pm 4 \sqrt{2}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \left(- \frac{6}{2}\right) \pm \frac{4 \sqrt{2}}{2}$
$x = \left(3 \pm 2 \sqrt{2}\right)$