# Question #c1f53

Jun 18, 2017

$x = 15 \pm 2 \sqrt{7}$

#### Explanation:

In order to solve this, you must be aware of something called the square root property:

We are given:

${\left(x - 15\right)}^{2} = 28$

The goal here is to isolate $x$ but we have to deal with the exponent. We want to get rid of the exponent and we can do this by taking the square root of the left side as well as the right side to keep the equation balanced:

$\sqrt{{\left(x - 15\right)}^{2}} = \sqrt{28}$

$x - 15 = \pm 2 \sqrt{7}$ *

*We can break down $\sqrt{28}$ into $\sqrt{4 \cdot 7}$ which simplifies to $\pm 2 \sqrt{7}$

From this point we just solve for $x$

Add $15$ to both sides:

$x \cancel{- 15 + 15} = 15 \pm 2 \sqrt{7}$

$x = 15 \pm 2 \sqrt{7}$