Answer
The solutions are $x=-3+\sqrt{2}$ or $x=-3-\sqrt{2}$
Work Step by Step
$ f(x)=x^{2}+6x+7\qquad$..an x-intercept is a point in the graph where the function intercepts the $\mathrm{x}$ axis.
If a function intercepts the $x$ axis, $f(x)=0$.
Substitute $0$ for $f(x)$ in the given function.
$ 0=x^{2}+6x+7\qquad$..add $-7$ to both sides so we can complete the square on the right side.
$-7=x^{2}+6x\qquad$..add $9$ to both sides to complete the square ($\displaystyle \frac{1}{2}(6)=3$, and $(3)^{2}=9$.)
$ x^{2}+6x+9=-7+9\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x+3)^{2}=2$
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+3=\pm\sqrt{2}\qquad$...add $-3$ to each side.
$ x+3-3=\pm\sqrt{2}-3\qquad$...simplify.
$x=-3\pm\sqrt{2}$
$x=-3+\sqrt{2}$ or $x=-3-\sqrt{2}$
