# How do you solve 5x^2-7=0?

Sep 25, 2015

$x = \pm \frac{\sqrt{35}}{5}$

#### Explanation:

You can solve this by isolating $x$.

5x^2−7=0
5x^2−7+7=0+7
$5 {x}^{2} = 7$
$\frac{1}{5} \left(5 {x}^{2}\right) = \frac{1}{5} \left(7\right)$
${x}^{2} = \frac{7}{5}$
$x = \pm \sqrt{\frac{7}{5}}$

You can have that as your final answer or you can rationalize it:
$x = \pm \sqrt{\frac{7}{5}}$
$x = \pm \sqrt{\frac{7}{5} \cdot \frac{5}{5}}$
$x = \pm \sqrt{\frac{7 \cdot 5}{5} ^ 2}$
$x = \pm \frac{\sqrt{35}}{5}$