## Algebra 2 (1st Edition)

The solutions are $r=-\displaystyle \frac{1}{2}+\frac{i\sqrt{7}}{2}$ and $r=-\displaystyle \frac{1}{2}-\frac{i\sqrt{7}}{2}$.
$6r^{2}+6r+12=0\qquad$ ...divide each term with $6$. $r^{2}+r+2=0\qquad$ ...add $-2$ to each side. $r^{2}+r+2-2=0-2\qquad$ ...simplify. $r^{2}+r=-2\qquad$ ...square half the coefficient of $r$. $(\displaystyle \frac{1}{2})^{2}=\frac{1}{4}\qquad$ ...add $\displaystyle \frac{1}{4}$ to each side of the expression $r^{2}+r+\displaystyle \frac{1}{4}=-2+\frac{1}{4}\qquad$ ...simplify. $r^{2}+r+\displaystyle \frac{1}{4}=-\frac{7}{4}\qquad$ ... write left side as a binomial squared. $(r+\displaystyle \frac{1}{2})^{2}=-\frac{7}{4}\qquad$ ...take square roots of each side. $r+\displaystyle \frac{1}{2}=\pm\sqrt{-\frac{7}{4}}\qquad$ ...rewrite $\sqrt{-\frac{7}{4}}$ as $\sqrt{-1\cdot\frac{1}{4}\cdot 7}$ $r+\displaystyle \frac{1}{2}=\pm\sqrt{-1\cdot\frac{1}{4}\cdot 7}\qquad$ ...simplify. $r+\displaystyle \frac{1}{2}=\pm\frac{i\sqrt{7}}{2}\qquad$ ...add $-\displaystyle \frac{1}{2}$ to each side. $r=-\displaystyle \frac{1}{2}\pm\frac{i\sqrt{7}}{2}$