We have, $x^2$ + 3x + 9 = 0 Comparing the given equation with the general form $ax^2$ + bx + c = 0, we get a = 1, b = 3, c = 9 ∴ x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ = $\frac{-3\pm\sqrt{(3)^2-4\times1\times9}}{2.1}$ = $\frac{-3\pm\sqrt{9-36}}{2}$ = $\frac{-3\pm\sqrt{-27}}{2}$ = $\frac{-3\pm3\sqrt{-3}}{2}$ = $\frac{-3\pm i(3\sqrt{3})}{2}$ ∴ The roots of the given equation are $\frac{-3+i(3\sqrt{3})}{2}$ and $\frac{-3-i(3\sqrt{3})}{2}$