#### Answer

The solutions are $x=\dfrac{3}{2}\pm\dfrac{\sqrt{3}}{6}i$

#### Work Step by Step

$-3x^{2}+9x=7$
Divide the whole equation by $-3$:
$-\dfrac{1}{3}(-3x^{2}+9x=7)$
$x^{2}-3x=-\dfrac{7}{3}$
Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-3$
$x^{2}-3x+\Big(-\dfrac{3}{2}\Big)^{2}=-\dfrac{7}{3}+\Big(-\dfrac{3}{2}\Big)^{2}$
$x^{2}-3x+\dfrac{9}{4}=-\dfrac{7}{3}+\dfrac{9}{4}$
$x^{2}-3x+\dfrac{9}{4}=-\dfrac{1}{12}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{3}{2}\Big)^{2}=-\dfrac{1}{12}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{3}{2}\Big)^{2}}=\pm\sqrt{-\dfrac{1}{12}}$
$x-\dfrac{3}{2}=\pm\dfrac{1}{2\sqrt{3}}i$
$x-\dfrac{3}{2}=\pm\dfrac{\sqrt{3}}{6}i$
Solve for $x$:
$x=\dfrac{3}{2}\pm\dfrac{\sqrt{3}}{6}i$