Answer
{$7 - 2\sqrt {17},7 + 2\sqrt {17}$}
Work Step by Step
Step 1: Comparing $n^{2}-14n-19=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=1$, $b=-14$ and $c=-19$
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:
$n=\frac{-(-14) \pm \sqrt {(-14)^{2}-4(1)(-19)}}{2(1)}$
Step 4: $n=\frac{14 \pm \sqrt {196+76}}{2}$
Step 5: $n=\frac{14 \pm \sqrt {272}}{2}$
Step 6: $n=\frac{14 \pm \sqrt {16\times17}}{2}$
Step 7: $n=\frac{14 \pm (\sqrt {16}\times\sqrt {17})}{2}$
Step 8: $n=\frac{14 \pm (4\times \sqrt {17})}{2}$
Step 9: $n=\frac{2(7 \pm 2\sqrt {17})}{2}$
Step 10: $n=7 - 2\sqrt {17}$ or $n=7 + 2\sqrt {17}$
Step 11: Therefore, the solution set is {$7 - 2\sqrt {17},7 + 2\sqrt {17}$}.
