Answer
{$-1 - \sqrt {10},-1 +\sqrt {10}$}
Work Step by Step
Step 1: Comparing $n^{2}+2n-9=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=1$, $b=2$ and $c=-9$
Step 2: The quadratic formula to be used is:
$n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$n=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(-9)}}{2(1)}$
Step 4: $n=\frac{-2 \pm \sqrt {4+36}}{2}$
Step 5: $n=\frac{-2 \pm \sqrt {40}}{2}$
Step 6: $n=\frac{-2 \pm \sqrt {4\times10}}{2}$
Step 7: $n=\frac{-2 \pm (\sqrt {4}\times\sqrt {10})}{2}$
Step 8: $n=\frac{-2 \pm (2\times \sqrt {10})}{2}$
Step 9: $n=\frac{2(-1 \pm 1\sqrt {10})}{2}$
Step 10: $n=-1 - 1\sqrt {10}$ or $n=-1 + 1\sqrt {10}$
Step 11: Therefore, the solution set is {$-1 - \sqrt {10},-1 +\sqrt {10}$}.
