Answer
$\frac{(x-3)^2}{25}+\frac{(y+2)^2}{4}=1$, foci: $(3\pm\sqrt {21},-2)$.
Work Step by Step
Step 1. Rewrite the given equation as
$4(x^2-6x+9)+25(y^2+4y+4)=36+100-36=100$ or $\frac{(x-3)^2}{25}+\frac{(y+2)^2}{4}=1$,
Step 2. We have $a^2=25, b^2=4$ and $c=\sqrt {a^2-b^2}=\sqrt {21}$. The ellipse is centered at $(3,-2)$ with a horizontal major axis.
Step 3. We can graph the equation as shown in the figure with foci at $(3\pm\sqrt {21},-2)$.
