Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The solutions are $x=-\displaystyle \frac{4}{3}$ and $x=-\displaystyle \frac{2}{3}$.
$9x^{2}+18x=-8\qquad$..divide both sides with $9$ $x^{2}+2x=-\displaystyle \frac{8}{9}\qquad$..add $1$ to both sides to complete the square ($\displaystyle \frac{1}{2}(2)=1$, and $(1)^{2}=1$.) $x^{2}+2x+1=-\displaystyle \frac{8}{9}+1\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+1)^{2}=\displaystyle \frac{1}{9}$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $x+1=\pm\sqrt{\frac{1}{9}}\qquad$...add $-1$ to each side. $x+1-1=\pm\sqrt{\frac{1}{9}}-1\qquad$...simplify. $x=-1\pm\sqrt{\frac{1}{9}}\qquad$...apply the quotient rule for radicals: $\displaystyle \sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}}$ and simplify $x=-1+\displaystyle \frac{1}{3}$ or $x=-1-\displaystyle \frac{1}{3}$ $x=-\displaystyle \frac{2}{3}$ or $x=-\displaystyle \frac{4}{3}$