# How do I find all real and complex zeros of x^2+3x+12?

Mar 2, 2015

First of all, let's find the discriminant:

$\Delta = {b}^{2} - 4 a c = 9 - 48 = - 39$.

Since $\Delta < 0$, the equation has two complex conjugate solution:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 3 \pm \sqrt{- 39}}{2} = \frac{- 3 \pm \sqrt{39} \sqrt{- 1}}{2} =$

$= - \frac{3}{2} \pm i \frac{\sqrt{39}}{2}$.

Remember that $\sqrt{- 1} = i$.