Answer
$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
Work Step by Step
The foci, center and vertices lie on the same line, the main axis.
Main axis here: $y=0$ (the x-axis).
Table 3: Major axis horizontal (parallel to the x-axis)$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\
{(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$
---
Center: $(0,0)$
Focus: $(3,0) \quad \Rightarrow \quad c=3$
Vertex: $(5,0) \quad \Rightarrow \quad a=5$
Find b:
$b^{2}=a^{2}-c^{2}=25-9=16$
$b=4$
The equation is $ \quad \displaystyle \frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$
or
$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$