Answer
$\left(-\frac{5}{2}-\frac{\sqrt{17}}{2}, 0\right)$ and $\left(-\frac{5}{2}+\frac{\sqrt{17}}{2}, 0\right)$
Work Step by Step
The $x$-intercepts of a function can be found by setting $y=0$ then solving for $x$.
Replace $g(x)$ by $0$, then solve for $x$ to obtain:
\begin{align*}
0&=x^2+5x+2\\
-25&=x^2+5x
\end{align*}
Complete the square by adding $\left(\dfrac{5}{2}\right)^2=\dfrac{25}{4}$ to both sides to obtain:
\begin{align*}
-2+\frac{25}{4}&=x^2+5x+\frac{25}{4}\\\\
-\frac{8}{4}+\frac{25}{4}&=\left(x+\frac{5}{2}\right)^2\\\\
\frac{17}{4}&=\left(x+\frac{5}{2}\right)^2\\\\
\pm\sqrt{\frac{17}{4}}&=\sqrt{\left(x+\frac{5}{2}\right)^2}\\\\
\pm\frac{\sqrt{17}}{2}&=x+\frac{5}{2}\\\\
-\frac{5}{2}\pm\frac{\sqrt{17}}{2}&=x
\end{align*}
Thus, the $x$-intercepts are:
$\left(-\frac{5}{2}-\frac{\sqrt{17}}{2}, 0\right)$ and $\left(-\frac{5}{2}+\frac{\sqrt{17}}{2}, 0\right)$