# How do you solve 5(x+3/4)^2-2=43?

Feb 7, 2017

$x = - \frac{15}{4} \text{ or } x = \frac{9}{4}$

#### Explanation:

$\text{Isolate } {\left(x + \frac{3}{4}\right)}^{2}$ on the left side of the equation.

$5 {\left(x + \frac{3}{4}\right)}^{2} \cancel{- 2} \cancel{+ 2} = 43 + 2$

$\Rightarrow 5 {\left(x + \frac{3}{4}\right)}^{2} = 45$

divide both sides by 5

$\frac{\cancel{5}}{\cancel{5}} {\left(x + \frac{3}{4}\right)}^{2} = \frac{45}{5}$

$\Rightarrow {\left(x + \frac{3}{4}\right)}^{2} = 9$

Take the $\textcolor{b l u e}{\text{square root }}$ of both sides.

$\sqrt{{\left(x + \frac{3}{4}\right)}^{2}} = \pm \sqrt{9}$

$\Rightarrow x + \frac{3}{4} = \pm 3 \leftarrow \textcolor{red}{\text{ 2 solutions}}$

$\textcolor{b l u e}{\text{Solution 1}}$

$x = - 3 - \frac{3}{4} \Rightarrow x = - \frac{15}{4}$

$\textcolor{b l u e}{\text{Solution 2}}$

$x = + 3 - \frac{3}{4} = \frac{9}{4}$