Answer
The solutions are $x=-9$ or $x=1$.
Work Step by Step
$ x^{2}+8x=9\qquad$..add $16$ to both sides to complete the square ($\displaystyle \frac{1}{2}(8)=4$, and $4^{2}=16$.)
$ x^{2}+8x+16=9+16\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x+4)^{2}=25$
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+4=\pm\sqrt{25}\qquad$...add $-4$ to each side.
$ x+4-4=\pm\sqrt{25}-4\qquad$...simplify.
$x=-4\pm 5$
$x=-4+5$ or $x=-4-5$
$x=1$ or $x=-9$
