Answer
center $(2,-1)$, foci $(1,-1),(3,-1)$, vertices $(2-\sqrt 3,-1),(2+\sqrt 3,-1)$
See graph.
Work Step by Step
1. $2x^2+3y^2-8x+6y+5=0\Longrightarrow 2(x-2)^2+3(y+1)^2=6$ or $\frac{(x-2)^2}{3}+\frac{(y+1)^2}{2}=1$,
2. we can find center $(2,-1)$ with a horizontal major axis, $a=\sqrt 3, b=\sqrt 2$, $c=\sqrt {a^2-b^2}=\sqrt {3-2}=1$,
3. thus foci $(2\pm1,-1)$ or $(1,-1),(3,-1)$, vertices $(2\pm\sqrt 3,-1)$ or $(2-\sqrt 3,-1),(2+\sqrt 3,-1)$
4. See graph.
