We have, $x^2$ + 3x + 5 = 0 Comparing the given equation with the general form $ax^2$ + bx + c = 0, we get a = 1, b = 3, c = 5. ∴ x = $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ = $\frac{-3 \pm \sqrt{(3)^2-4\times1\times5}}{2}$ = $\frac{-3 \pm \sqrt{9-20}}{2}$ = $\frac{-3 \pm \sqrt{-11}}{2}$ = $\frac{-3 \pm \sqrt{11}}{2}$ ∴ The roots of the given equation are $\frac{-3 + \sqrt{-11}}{2}$ and $\frac{-3 - \sqrt{-11}}{2}$