Answer
{$-1-i\sqrt 6,-1+i\sqrt 6$}
Work Step by Step
Step 1: Comparing $n^{2}+2n+7=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=1$, $b=2$ and $c=7$
Step 2: The quadratic formula is:
$n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$n=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(7)}}{2(1)}$
Step 4: $n=\frac{-2 \pm \sqrt {4-28}}{2}$
Step 5: $n=\frac{-2 \pm \sqrt {-24}}{2}$
Step 6: $n=\frac{-2 \pm \sqrt {-1\times24}}{2}$
Step 7: $n=\frac{-2 \pm (\sqrt {-1}\times\sqrt {24})}{2}$
Step 8: $n=\frac{-2 \pm (i\times \sqrt {4\times6})}{2}$
Step 9: $n=\frac{-2 \pm i(2\sqrt 6)}{2}$
Step 10: $n=\frac{2(-1 \pm i\sqrt 6)}{2}$
Step 11: $n=-1 \pm i\sqrt 6$
Step 12: $n=-1-i\sqrt 6$ or $x=-1+i\sqrt 6$
Step 13: Therefore, the solution set is {$-1-i\sqrt 6,-1+i\sqrt 6$}.
