Answer
$ \displaystyle \frac{(x-2)^{2}}{3}+\frac{(y-3)^{2}}{2}=1$
New foci:$\quad (3,3)$ and $(1, 3)$
New vertices:$\quad (2\pm\sqrt{3},3)$
New center:$\quad (2,3)$
Work Step by Step
Shift right by $2 \Rightarrow$ in the equation, replace $x$ with $x-2$
Shift up by $3 \Rightarrow$ in the equation, replace $y$ with $y-3$
New equation:$\displaystyle \quad \frac{(x-2)^{2}}{3}+\frac{(y-3)^{2}}{2}=1$
The ellipse $\displaystyle \frac{x^{2}}{3}+\frac{y^{2}}{2}=1$
Foci on the x-axis: $\quad \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad (a\gt b)$
$a=\sqrt{3}\quad b=\sqrt{2}$
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}=\sqrt{3-2}=1$
Foci: $\quad(\pm 1, 0)$
Vertices: $\quad (\pm\sqrt{3}, 0)$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x+2 & \text{... shift right}\\
y'=y+3 & \text{... shift up}
\end{array}\right.$
New foci:
$\quad (\pm 1+2, 0+3)\Rightarrow\quad (3,3)$ and $(1, 3)$
New vertices:
$\quad (\pm\sqrt{3}+2, 0+3)\Rightarrow\quad (2\pm\sqrt{3},3)$
New center:
$\quad (0+2,0+3)\quad =(2,3)$
