# How do you solve -(3/4)x^2 + 2 = 0 ?

Aug 21, 2015

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

#### Explanation:

Start by isolating ${x}^{2}$ on one side of the equation. You can do that by adding $- 2$ to both sides first

$- \frac{3}{4} {x}^{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = - 2$

$- \frac{3}{4} {x}^{2} = - 2$

then multiply both sides by $- \frac{4}{3}$ to get

$\left(- \frac{4}{3}\right) \cdot \left(- \frac{3}{4}\right) {x}^{2} = - 2 \cdot \left(- \frac{4}{3}\right)$

${x}^{2} = \frac{8}{3}$

To solve for $x$, take the square root of both sides

$\sqrt{{x}^{2}} = \sqrt{\frac{8}{3}}$

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

You can simplify this by rationalizing the denominator

$x = \pm \frac{2 \sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \textcolor{g r e e n}{\pm \frac{2 \sqrt{6}}{3}}$