Quadratics by Completing the Square (Complex Roots)

For $x^2 + px + q = 0$, completing the square gives:

$\left(x + \frac{p}{2}\right)^2 = \left(\frac{p}{2}\right)^2 - q$

When the right-hand side is negative — say it equals $-D$ for some positive $D$ — the roots are complex:

$x = -\frac{p}{2} \pm i\sqrt{D}$

Example: $x^2 + 2x + 5 = 0$. Here $(x+1)^2 = 1-5 = -4$, so $x + 1 = \pm 2i$ and $x = -1 \pm 2i$.