Answer
The solutions are $\displaystyle \frac{1}{6}+i\frac{\sqrt{71}}{6}$ and $ \displaystyle \frac{1}{6}-i\frac{\sqrt{71}}{6}$
Work Step by Step
$ 3x^{2}+x=2x-6\qquad$ ...add $-2x$ to each side
$ 3x^{2}-x=-6\qquad$ ...divide the entire expression with $3$.
$ x^{2}-\displaystyle \frac{1}{3}x=-2\qquad$ ...square half the coefficient of $q$.
$(\displaystyle \frac{\frac{-1}{3}}{2})^{2}=(\frac{1}{6})^{2}\qquad$ ...complete the square by adding$ \displaystyle \frac{1}{36}$ to each side of the expression
$ x^{2}-\displaystyle \frac{1}{3}x+\frac{1}{36}=-2+\frac{1}{36}\qquad$ ...Write $x^{2}-\displaystyle \frac{1}{3}x+\frac{1}{36}$ as a binomial squared.
$(x-\displaystyle \frac{1}{6})^{2}=-\frac{71}{36}\qquad$ ...take square roots of each side.
$ x-\displaystyle \frac{1}{6}=\pm\sqrt{-\frac{71}{36}}\qquad$ ...simplify $\displaystyle \sqrt{-\frac{71}{36}}=i\sqrt{\frac{71}{36}}=i\frac{\sqrt{71}}{\sqrt{36}}=i\frac{\sqrt{71}}{6}$
$ x-\displaystyle \frac{1}{6}=\pm i\frac{\sqrt{71}}{6}\qquad$ ...add $\displaystyle \frac{1}{6}$ to each side.
$x=\displaystyle \frac{1}{6}\pm i\frac{\sqrt{71}}{6}$
