Answer
$\{-2\pm i\}$
Work Step by Step
$x^{2}+4x+5=0$
$a=1,b=4,c=5$
$b^{2}-4ac=16-20=-4$, negative
so the solutions are a complex conjugate pair
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-4\pm\sqrt{-4}}{2}$
$=\displaystyle \frac{-4\pm 2i}{2}$
$=\displaystyle \frac{2(-2\pm i)}{2}=-2\pm i$
Solution set = $\{-2\pm i\}$
