# How do you solve (x-3)^2=36?

##### 2 Answers
Oct 31, 2015

$x = 9 \mathmr{and} x = - 3$

#### Explanation:

${\left(x - 3\right)}^{2} = 36$

${\left(x - 3\right)}^{2} = {6}^{2}$

$\sqrt{{\left(x - 3\right)}^{2}}$ = 6

$\pm \left(x - 3\right)$= 6

either,
taking +,
$+ \left(x - 3\right) = 6$
$x - 3 = 6$
$x = 6 + 3$
So,
$x = 9$

Or,
Taking -,
$- \left(x - 3\right) = 6$
$- x + 3 = 6$
$- x = 6 - 3$
$- x = 3$
So,
$x = - 3$

Oct 31, 2015

I found:
${x}_{1} = 9$
${x}_{2} = - 3$

#### Explanation:

You can take the square root of both sides and get:
$\left(x - 3\right) = \pm \sqrt{36}$
$x = 3 \pm 6$
so you have two possibilities:
${x}_{1} = 3 + 6 = 9$
${x}_{2} = 3 - 6 = - 3$
Both work when inserted into your original equation.