Answer
{$\frac{-4-\sqrt {11}}{5},\frac{-4+\sqrt {11}}{5}$}
Work Step by Step
Step 1: Comparing $5n^{2}+8n+1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=5$, $b=8$ 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:
$n=\frac{-(8) \pm \sqrt {(8)^{2}-4(5)(1)}}{2(5)}$
Step 4: $n=\frac{-8 \pm \sqrt {64-20}}{10}$
Step 5: $n=\frac{-8 \pm \sqrt {44}}{10}$
Step 6: $n=\frac{-8 \pm \sqrt {4\times11}}{10}$
Step 7: $n=\frac{-8 \pm 2\sqrt {11}}{10}$
Step 8: $n=\frac{-4 \pm \sqrt {11}}{5}$
Step 9: $n=\frac{-4-\sqrt {11}}{5}$ or $n=\frac{-4+\sqrt {11}}{5}$
Step 10: Therefore, the solution set is {$\frac{-4-\sqrt {11}}{5},\frac{-4+\sqrt {11}}{5}$}.
