Answer
{$\frac{-2-\sqrt {5}}{2},\frac{-2+\sqrt {5}}{2}$}
Work Step by Step
Step 1: Comparing $4n^{2}+8n-1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=4$, $b=8$ and $c=-1$
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:
$n=\frac{-(8) \pm \sqrt {(8)^{2}-4(4)(-1)}}{2(4)}$
Step 4: $n=\frac{-8 \pm \sqrt {64+16}}{8}$
Step 5: $n=\frac{-8 \pm \sqrt {80}}{8}$
Step 6: $n=\frac{-8 \pm \sqrt {16\times5}}{8}$
Step 7: $n=\frac{-8 \pm 4\sqrt {5}}{8}$
Step 8: $n=\frac{4(-2 \pm \sqrt {5})}{8}$
Step 9: $n=\frac{(-2 \pm \sqrt {5})}{2}$
Step 10: $n=\frac{-2-\sqrt {5}}{2}$ or $n=\frac{-2+\sqrt {5}}{2}$
Step 11: Therefore, the solution set is {$\frac{-2-\sqrt {5}}{2},\frac{-2+\sqrt {5}}{2}$}.
