## Algebra 2 (1st Edition)

The solutions are $-4$ and $-5$.
$x^{2}+9x+20=0\qquad$ ...Write left side in the form $x^{2}+bx.$(add $-20$ to each side) $x^{2}+9x=-20\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{9}{2})^{2}=\frac{81}{4}\qquad$ ...complete the square by adding$\displaystyle \frac{81}{4}$ to each side of the expression $x^{2}+9x+\displaystyle \frac{81}{4}=-20+\frac{81}{4}\qquad$ ...Write left side as a binomial squared. $(x+\displaystyle \frac{9}{2})^{2}=\frac{1}{4}\qquad$ ...take square roots of each side. $x+\displaystyle \frac{9}{2}=\pm\sqrt{\frac{1}{4}}\qquad$ ...evaluate$\sqrt{\frac{1}{4}}$ $x+\displaystyle \frac{9}{2}=\pm\frac{1}{2}\qquad$ ...add $-\displaystyle \frac{9}{2}$ to each side. $x=-\displaystyle \frac{9}{2}\pm\frac{1}{2}$