Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 467: 6

{$11,15$}

Using the rules for the factoring of trinomials, we obtain: $n^{2}-26n+165=0$ $n^{2}-11n-15n+165=0$ $n(n-11)-15(n-11)=0$ $(n-11)(n-15)=0$ $(n-11)=0$ and $(n-15)=0$ $n=11$ and $n=15$ Therefore, the solution set is {$11,15$}.