Answer
The x-intercepts are $(-5+\sqrt{27},0)$ and $(-5-\sqrt{27},0)$
Work Step by Step
$ f(x)=x^{2}+10x-2\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-2\qquad$..add $2$ to both sides so we can complete the square.
$ 2=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=2+25\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x+5)^{2}=27$
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{27}\qquad$...add $-5$ to each side.
$ x+5-5=\pm\sqrt{27}-5\qquad$...simplify.
$x=-5\pm\sqrt{27}$
$x=-5+\sqrt{27}$ or $x=-5-\sqrt{27}$
