Answer
{$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}
Work Step by Step
Step 1: Comparing $4n^{2}+6n-1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=4$, $b=6$ 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:
$n=\frac{-(6) \pm \sqrt {(6)^{2}-4(4)(-1)}}{2(4)}$
Step 4: $n=\frac{-6 \pm \sqrt {36+16}}{8}$
Step 5: $n=\frac{-6 \pm \sqrt {52}}{8}$
Step 6: $n=\frac{-6 \pm \sqrt {4\times13}}{8}$
Step 7: $n=\frac{-6 \pm 2\sqrt {13}}{8}$
Step 8: $n=\frac{2(-3 \pm \sqrt {13})}{8}$
Step 9: $n=\frac{(-3 \pm \sqrt {13})}{4}$
Step 10: $n=\frac{-3-\sqrt {13}}{4}$ or $n=\frac{-3+\sqrt {13}}{4}$
Step 11: Therefore, the solution set is {$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}.
