Answer
The solutions are $r=-\displaystyle \frac{1}{2}+\frac{i\sqrt{7}}{2}$ and $r=-\displaystyle \frac{1}{2}-\frac{i\sqrt{7}}{2}$.
Work Step by Step
$ 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.
$r=-\displaystyle \frac{1}{2}\pm\frac{i\sqrt{7}}{2}$
