Answer
{$\frac{-9-\sqrt {57}}{12},\frac{-9+\sqrt {57}}{12}$}
Work Step by Step
Step 1: Comparing $6n^{2}+9n+1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=6$, $b=9$ 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:
$n=\frac{-(9) \pm \sqrt {(9)^{2}-4(6)(1)}}{2(6)}$
Step 4: $n=\frac{-9 \pm \sqrt {81-24}}{12}$
Step 5: $n=\frac{-9 \pm \sqrt {57}}{12}$
Step 6: $n=\frac{-9-\sqrt {57}}{12}$ or $n=\frac{-9+\sqrt {57}}{12}$
Step 7: Therefore, the solution set is {$\frac{-9-\sqrt {57}}{12},\frac{-9+\sqrt {57}}{12}$}.
