Answer
The x-intercepts are $(4+\sqrt{26},0)$ and $(4-\sqrt{26},0)$
Work Step by Step
$ f(x)=x^{2}-8x-10\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}-8x-10\qquad$..add $10$ to both sides so we can complete the square.
$ 10=x^{2}-8x\qquad$..add $16$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-8)=-4$, and $(-4)^{2}=16$.)
$ x^{2}-8x+16=10+16\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x-4)^{2}=26$
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{26}\qquad$...add $4$ to each side.
$ x-4+4=\pm\sqrt{26}+4\qquad$...simplify.
$x=4\pm\sqrt{26}$
$x=4+\sqrt{26}$ or $x=4-\sqrt{26}$
