Answer
{$-15,13$}
Work Step by Step
Step 1: Comparing $n^{2}+2n-195=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=2$, and $c=-195$
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, we obtain:
$x=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(-195)}}{2(1)}$
Step 4: $x=\frac{-2 \pm \sqrt {4+780}}{2}$
Step 5: $x=\frac{-2 \pm \sqrt {784}}{2}$
Step 6: $x=\frac{-2 \pm 28}{2}$
Step 7: $x=\frac{-2+28}{2}$ or $x=\frac{-2-28}{2}$
Step 8: $x=\frac{26}{2}$ or $x=\frac{-30}{2}$
Step 9: $x=13$ or $x=-15$
Step 10: Therefore, the solution set is {$-15,13$}.
