Answer
{$2-4i,2+4i$}
Work Step by Step
Step 1: Comparing $x^{2}-4x+20=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=-4$ and $c=20$
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:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(20)}}{2(1)}$
Step 4: $x=\frac{4 \pm \sqrt {16-80}}{2}$
Step 5: $x=\frac{4 \pm \sqrt {-64}}{2}$
Step 6: $x=\frac{4 \pm \sqrt {-1\times64}}{2}$
Step 7: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {64})}{2}$
Step 8: $x=\frac{4 \pm (i\times 8)}{2}$
Step 9: $x=\frac{4 \pm i(8)}{2}$
Step 10: $x=\frac{2(2 \pm 4i)}{2}$
Step 11: $x=2 \pm 4i$
Step 12: $x=2+4i$ or $x=2-4i$
Step 13: Therefore, the solution set is {$2-4i,2+4i$}.
