## Precalculus (6th Edition)

$\color{blue}{\left\{-\frac{5}{2}, 2\right\}}$
Divide $28$ to both sides: $x^2+\frac{1}{2}x=5$ Add the square of one-half of the coefficient of $x$, which is $(\dfrac{\frac{1}{2}}{2})^2=(\frac{1}{4})^2=\frac{1}{16}$, to obtain: $x^2+\frac{1}{2}x+\frac{1}{16}=5+\frac{1}{16} \\(x+\frac{1}{4})^2=\frac{80}{16}+\frac{1}{16} \\(x+\frac{1}{4})^2=\frac{81}{16}$ Take the square root of both sides to obtain: $\sqrt{(x+\frac{1}{4})^2}=\pm \sqrt{\frac{81}{16}} \\x+\frac{1}{4} = \pm \frac{9}{4}$ Add $-\frac{1}{4}$ to both sides: $x=-\frac{1}{4} \pm \frac{9}{4} \\x_1=-\frac{1}{4}-\frac{9}{4}=-\frac{10}{4}=-\frac{5}{2} \\x_2=-\frac{1}{4}+\frac{9}{4}=\frac{8}{4}=2$ Thus, the solution set is: $\color{blue}{\left\{-\frac{5}{2}, 2\right\}}$.