# How do you solve r^2+25=0 using the quadratic formula?

Dec 3, 2016

$r = - 5 i \mathmr{and} r = + 5 i$

#### Explanation:

$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$
Converting the given: ${r}^{2} + 25 = 0$ into standard form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{1} {r}^{2} + \textcolor{b l u e}{0} r + \textcolor{g r e e n}{25} = 0$
$\textcolor{w h i t e}{\text{XXX}} r = \frac{\textcolor{b l u e}{0} \pm \sqrt{{\textcolor{b l u e}{0}}^{2} - 4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{25}}}{2 \cdot \textcolor{red}{1}}$
$\textcolor{w h i t e}{\text{XXX}} = \pm \frac{2 \sqrt{- 25}}{2} = \pm \sqrt{- 25} = \pm 5 i$