# How do you solve r^2+3r-3=0 by completing the square?

Jul 26, 2015

I found:
${r}_{1} = \frac{- 3 + \sqrt{21}}{2}$
${r}_{2} = \frac{- 3 - \sqrt{21}}{2}$

#### Explanation:

Write it as:
${r}^{2} + 3 r = 3$
add and subtract $\frac{9}{4}$ (i.e., a number that comes from $\frac{3}{2}$ so that squared gives you $\frac{9}{4}$ and multiplied by $2$ gives you $3$ which takes care of $3 r$) to get:
${r}^{2} + 3 r \textcolor{red}{+ \frac{9}{4} - \frac{9}{4}} = 3$ rearranging:
${r}^{2} + 3 r + \frac{9}{4} = 3 + \frac{9}{4}$
${\left(r + \frac{3}{2}\right)}^{2} = \frac{12 + 9}{4}$
${\left(r + \frac{3}{2}\right)}^{2} = \frac{21}{4}$ square root both sides:
$r + \frac{3}{2} = \pm \sqrt{\frac{21}{4}}$ you get two solutions:
${r}_{1} = \frac{- 3 + \sqrt{21}}{2}$
${r}_{2} = \frac{- 3 - \sqrt{21}}{2}$