Answer
$\dfrac{3-3i\sqrt3}{2}, \dfrac{3+3i\sqrt3}{2}, 3$
Work Step by Step
Use the Zero Factor Property by equating each factor to zero, then solve each equation:
\begin{array}{ccc}
&x+3=0 &\text{or} &x^2-3x+9=0
\\&x=-3 & \text{or} &x^2-3x+9=0
\end{array}
Solve the second equation using the quadratic formula with $a=1, b=-3, c=9$ to obtain:
$x=\dfrac{-(-3)\pm \sqrt{(-3)^2-4(1)(9)}}{2(1)}
\\x=\dfrac{3\pm \sqrt{9-36}}{2}
\\x=\dfrac{3\pm \sqrt{-27}}{2}
\\x=\dfrac{3 \pm \sqrt{9(-3)}}{2}
\\x=\dfrac{3 \pm 3\sqrt{-3}}{2}
\\x=\dfrac{3\pm 3i\sqrt{3}}{2}$
Therefore, the solutions of the equation are:
$\dfrac{3-3i\sqrt3}{2}, \dfrac{3+3i\sqrt3}{2}, 3$
