Answer
{$2-i,2+i$}
Work Step by Step
Step 1: Comparing $n^{2}-4n+5=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=1$, $b=-4$ 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:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(5)}}{2(1)}$
Step 4: $x=\frac{4 \pm \sqrt {16-20}}{2}$
Step 5: $x=\frac{4 \pm \sqrt {-4}}{2}$
Step 6: $x=\frac{4 \pm \sqrt {-1\times4}}{2}$
Step 7: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {2\times2})}{2}$
Step 8: $x=\frac{4 \pm (i\times 2)}{2}$
Step 9: $x=\frac{4 \pm 2i}{2}$
Step 10: $x=\frac{2(2 \pm i)}{2}$
Step 11: $x=2 \pm i$
Step 12: $x=2-i$ or $x=2+i$
Step 13: Therefore, the solution set is {$2-i,2+i$}.
