Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 467: 17

{$20,24$}

Using the rules for the factoring of trinomials, we obtain: $n^{2}-44n+480=0$ $n^{2}-20n-24n+480=0$ $n(n-20)-24(n-20)=0$ $(n-20)(n-24)=0$ $(n-20)=0$ and $(n-24)=0$ $n=20$ and $n=24$ Therefore, the solution set is {$20,24$}.