Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 13

{$2,12$}

Since, $u^2-4u-21=0$ Factorize the expression as follows: $(u-7)(u+3)=0$ or, $u=${$-3,7$} Replace $u$ with $x-5$, we have $x-5=-3 \implies x=2$ and $x-5=7$ or, $x=12$ Hence, our solution set is $x=${$2,12$}