#### Answer

The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{3}\pm\frac{\sqrt6}{3}i\Big\}$$

#### Work Step by Step

$$3(3x^2-2x)=-7$$
The equation is not in standard form, so first, we need to bring it back to standard form:
$$9x^2-6x=-7$$
$$9x^2-6x+7=0$$
Now the equation is in standard form, the quadratic formula can be applied.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
As $a=9, b=-6, c=7$
$$x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times9\times7}}{2\times9}$$
$$x=\frac{6\pm\sqrt{36-252}}{18}$$
$$x=\frac{6\pm\sqrt{-216}}{18}$$
Now we rewrite $\sqrt{-216}=i\sqrt{216}=i\sqrt{36\times6}=6i\sqrt6$
$$x=\frac{6\pm 6i\sqrt6}{18}$$
Finally, we simplify
$$x=\frac{1\pm i\sqrt6}{3}$$
The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{3}\pm\frac{\sqrt6}{3}i\Big\}$$