Answer
{$-1-2i,-1+2i$}
Work Step by Step
Step 1: Comparing $x^{2}+2x+5=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=2$ and $c=5$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(5)}}{2(1)}$
Step 4: $x=\frac{-2 \pm \sqrt {4-20}}{2}$
Step 5: $x=\frac{-2 \pm \sqrt {-16}}{2}$
Step 6: $x=\frac{-2 \pm \sqrt {-1\times16}}{2}$
Step 7: $x=\frac{-2 \pm (\sqrt {-1}\times\sqrt {16})}{2}$
Step 8: $x=\frac{-2 \pm (i\times 4)}{2}$
Step 9: $x=\frac{-2 \pm 4i}{2}$
Step 10: $x=\frac{2(-1 \pm 2i)}{2}$
Step 11: $x=-1 \pm 2i$
Step 12: $x=-1-2i$ or $x=-1+2i$
Step 13: Therefore, the solution set is {$-1-2i,-1+2i$}.
