Answer
The x-intercepts are $(5+\sqrt{47},0)$ and $(-5-\sqrt{47},0)$
Work Step by Step
$ f(x)=x^{2}-10x-22\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}-10x-22\qquad$..add $22$ to both sides so we can complete the square.
$ 22=x^{2}-10x\qquad$..add $25$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-10)=-5$, and $(-5)^{2}=25$.)
$ x^{2}-10x+25=22+25\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x-5)^{2}=47$
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-5=\pm\sqrt{47}\qquad$...add $5$ to each side.
$ x-5+5=\pm\sqrt{47}+5\qquad$...simplify.
$x=5\pm\sqrt{47}$
$x=5+\sqrt{47}$ or $x=-5-\sqrt{47}$
