Answer
{$3-2i,3+2i$}
Work Step by Step
Step 1: Comparing $n^{2}-6n+13=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=1$, $b=-6$ and $c=13$
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:
$x=\frac{-(-6) \pm \sqrt {(-6)^{2}-4(1)(13)}}{2(1)}$
Step 4: $x=\frac{6 \pm \sqrt {36-52}}{2}$
Step 5: $x=\frac{6 \pm \sqrt {-16}}{2}$
Step 6: $x=\frac{6 \pm \sqrt {-1\times16}}{2}$
Step 7: $x=\frac{6 \pm (\sqrt {-1}\times\sqrt {4\times4})}{2}$
Step 8: $x=\frac{6 \pm (i\times 4)}{2}$
Step 9: $x=\frac{6 \pm 4i}{2}$
Step 10: $x=\frac{2(3 \pm 2i)}{2}$
Step 11: $x=3 \pm 2i$
Step 12: $x=3-2i$ or $x=3+2i$
Step 13: Therefore, the solution set is {$3-2i,3+2i$}.
