Answer
{$-14,-9$}
Work Step by Step
Step 1: Comparing $n^{2}+23n+126=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=23$, and $c=126$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a,b, and c in the formula, we obtain:
$x=\frac{-(23) \pm \sqrt {(23)^{2}-4(1)(126)}}{2(1)}$
Step 4: $x=\frac{-23 \pm \sqrt {529-504}}{2}$
Step 5: $x=\frac{-23 \pm \sqrt {25}}{2}$
Step 6: $x=\frac{-23 \pm 5}{2}$
Step 7: $x=\frac{-23+5}{2}$ or $x=\frac{-23-5}{2}$
Step 8: $x=\frac{-18}{2}$ or $x=\frac{-28}{2}$
Step 9: $x=-9$ or $x=-14$
Step 10: Therefore, the solution set is {$-14,-9$}.
