## Intermediate Algebra (6th Edition)

$\dfrac{3-3i\sqrt3}{2}, \dfrac{3+3i\sqrt3}{2}, 3$
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: $\dfrac{3-3i\sqrt3}{2}, \dfrac{3+3i\sqrt3}{2}, 3$