# How do you solve 2(x-3)^2=8?

Jul 11, 2017

$x = 5$ or $x = 1$

See below

#### Explanation:

2(x−3)^2=8

Divide both sides by 2: ${\left(x - 3\right)}^{2} = 4$

Take the square root of each side:

$\left(x - 3\right) = \pm 2$

$x - 3 = + 2 \text{ } \rightarrow x = 5$
or
$x - 3 = - 2 \text{ } \rightarrow x = 1$

$x = 5$ or $x = 1$

Jul 11, 2017

The answer is x= 5 and 1

#### Explanation:

First you start by evualting the $2 {\left(x - 3\right)}^{2}$ part of the equation.
$2 {\left(x - 3\right)}^{2}$=8
$2 \left({x}^{2} - 6 x + 9\right) = 8$
$2 {x}^{2} - 12 x + 18 = 8$
Then you move over the constant on the right side to the left .
$2 {x}^{2} - 12 x + 18 - 8 = 0$
$2 {x}^{2} - 12 x + 10 = 0$
Then divide the whole equation by 2.
$\frac{2 {x}^{2} - 12 x + 10}{2} = \frac{0}{2}$
${x}^{2} - 6 x + 5 = 0$
Then factor the equation normally.
$\left(x - 5\right) \left(x - 1\right) = 0$