# How do you solve 2(x-5)^2=3?

Aug 21, 2015

$x = 5 \pm \frac{\sqrt{6}}{2}$

#### Explanation:

The first thing you need to do is isolate ${\left(x - 5\right)}^{2}$ on one side of the equation by dividing both sides by $2$.

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot {\left(x - 5\right)}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{3}{2}$

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

Now you need to take the square root of both sides

$\sqrt{{\left(x - 5\right)}^{2}} = \sqrt{\frac{3}{2}}$

$x - 5 = \pm \frac{\sqrt{3}}{\sqrt{2}}$

Rationalize the denominator by multiplying the fraction by $1 = \frac{\sqrt{2}}{\sqrt{2}}$ to get

$x - 5 = \pm \frac{\sqrt{3} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \pm \frac{\sqrt{6}}{2}$

Finally, isolate $x$ on one side by adding $5$ to both sides of the equation

$x - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} = 5 \pm \frac{\sqrt{6}}{2}$

${x}_{1 , 2} = 5 \pm \frac{\sqrt{6}}{2}$

This equation will thus have two solutions,

${x}_{1} = \textcolor{g r e e n}{5 + \frac{\sqrt{6}}{2}} \text{ }$ or $\text{ } {x}_{2} = \textcolor{g r e e n}{5 - \frac{\sqrt{6}}{2}}$