Answer
$\frac{(x-3)^2}{16}+\frac{(y+2)^2}{25}=1$
Work Step by Step
1. Use the standard form $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
2. The center $(3,-2)$ gives $h=3, k=-2$
3. We have $b=\sqrt {a^2-c^2}=4$
4. Thus the equation is $\frac{(x-3)^2}{16}+\frac{(y+2)^2}{25}=1$
