Answer
The solutions are $x=\displaystyle \frac{2+i\sqrt{3}}{3}$ or $x=\displaystyle \frac{2-i\sqrt{3}}{3}$.
Work Step by Step
$ 9x^{2}-12x+4=-3\qquad$ ...write left side as a binomial squared.($(a\pm b)^{2}=a^{2}\pm 2ab+b^{2}$)
$a=3x,b=2$
$(3x-2)^{2}=-3\qquad$ ...take square roots of each side.
$ 3x-2=\pm\sqrt{-3}\qquad$ ...add $2$ to each side.
$ 3x-2+2=\pm\sqrt{-3}+2\qquad$ ...simplify.($\sqrt{-3}=i\sqrt{3}$).
$ 3x=2\pm i\sqrt{3}\qquad$ ...divide both sides wirh $2$.
$ x=\displaystyle \frac{2\pm i\sqrt{3}}{3}\qquad$ ...write as separate equations.
$x=\displaystyle \frac{2+i\sqrt{3}}{3}$ or $x=\displaystyle \frac{2-i\sqrt{3}}{3}$
