# How do you solve the quadratic 3r^2-5=-10 using any method?

Dec 2, 2016

$r = \pm \frac{\sqrt{15}}{3} i$

#### Explanation:

$3 {r}^{2} - 5 = - 10$
$\textcolor{w h i t e}{{a}^{2} a} + 5 \textcolor{w h i t e}{a {a}^{2} a} + 5 \textcolor{w h i t e}{a a a}$Add 5 to both sides

$3 {r}^{2} = - 5$

$\frac{3 {r}^{2}}{3} = \frac{- 5}{3} \textcolor{w h i t e}{a a a}$Divide both sides by 3

${r}^{2} = - \frac{5}{3}$

$\sqrt{{r}^{2}} = \sqrt{- \frac{5}{3}} \textcolor{w h i t e}{a a a}$Square root both sides

$r = \pm i \sqrt{\frac{5}{3}} \textcolor{w h i t e}{a a}$The negative inside the square root comes out as $i$.

$r = \pm i \sqrt{\frac{5}{3}} \cdot \sqrt{\frac{3}{3}} \textcolor{w h i t e}{a a a}$Rationalize the denominator

$r = \pm i \frac{\sqrt{15}}{3} = \pm \frac{\sqrt{15}}{3} i$