Answer
The solutions are $x=3$ or $x=-7$.
Work Step by Step
$ x^{2}+4x=21\qquad$..add $4$ to both sides to complete the square ($\displaystyle \frac{1}{2}(4)=2$, and $2^{2}=4$.)
$ x^{2}+4x+4=21+4\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x+2)^{2}=25$
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+2=\pm\sqrt{25}\qquad$...add $-2$ to each side.
$ x+2-2=\pm\sqrt{25}-2\qquad$...simplify.
$x=-2\pm 5$
$x=-2+5$ or $x=-2-5$
$x=3$ or $x=-7$
The solutions are $x=3$ or $x=-7$.
