Answer
$\left(-\frac{9}{2}-\frac{\sqrt{181}}{2}, 0\right)$ and $\left(-\frac{9}{2}+\frac{\sqrt{181}}{2}, 0\right)$
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+9x-25\\
25&=x^2+9x
\end{align*}
Complete the square by adding $\left(\dfrac{9}{2}\right)^2=\dfrac{81}{4}$ to both sides to obtain:
\begin{align*}
25+\frac{81}{4}&=x^2+9x+\frac{81}{4}\\\\
\frac{100}{4}+\frac{81}{4}&=\left(x+\frac{9}{2}\right)^2\\\\
\frac{181}{4}&=\left(x+\frac{9}{2}\right)^2\\\\
\pm\sqrt{\frac{181}{4}}&=\sqrt{\left(x+\frac{9}{2}\right)^2}\\\\
\pm\frac{\sqrt{181}}{2}&=x+\frac{9}{2}\\\\
-\frac{9}{2}\pm\frac{\sqrt{181}}{2}&=x
\end{align*}
Thus, the $x$-intercepts are:
$\left(-\frac{9}{2}-\frac{\sqrt{181}}{2}, 0\right)$ and $\left(-\frac{9}{2}+\frac{\sqrt{181}}{2}, 0\right)$
