Answer
The solutions are $x=-\displaystyle \frac{4}{3}$ and $x=-\displaystyle \frac{2}{3}$.
Work Step by Step
$ 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 $x=-\displaystyle \frac{4}{3}$
