jmc

The sum of the distances from $(0,0)$ to the two foci is $2+3=5.$ By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci must also be $5.$ So, in particular, if $(x,0)$ is the other $x$-intercept, the distance formula gives $$|x-3|+\sqrt{x^2+4}=5.$$Drawing out the ellipse, we see that $x>3,$ so we can drop the absolute values around $x-3.$ Then, solving for $x$, we have $$\begin{array}{rl}\sqrt{x^2+4}& =8-x\\ x^2+4& =x^2-16x+64\\ 16x& =60,\end{array}$$so $x=\frac{60}{16}=\frac{15}{4}.$ Thus the answer is $\boxed{\left(\frac{15}{4},0\right)}.$