Pentagon $ABCDE$ is inscribed in a circle such that $ACDE$ is a square with area $12$. What is the largest possible area of pentagon $ABCDE$?

$\text{(A) }9+3\sqrt{2}\qquad\text{(B) }13\qquad\text{(C) }12+\sqrt{2}\qquad\text{(D) }14\qquad\text{(E) }12+\sqrt{6}-\sqrt{3}$
1. **Understanding the problem**: We are given a pentagon $ABCDE$ inscribed in a circle such that $ACDE$ forms a square with an area of $12$. We need to find the largest possible area of the pentagon $ABCDE$.

2. **Finding the side length of the square**: 
   - The area of the square $ACDE$ is given as $12$.
   - Let the side length of the square be $s$. Then, $s^2 = 12$.
   - Solving for $s$, we get:
     \[
     s = \sqrt{12} = 2\sqrt{3}
     \]

3. **Finding the radius of the circumscribed circle**:
   - The diagonal of the square $ACDE$ is $s\sqrt{2}$.
   - Since the square is inscribed in the circle, the diagonal of the square is also the diameter of the circle.
   - Therefore, the diameter of the circle is $2r = s\sqrt{2}$.
   - Substituting $s = 2\sqrt{3}$, we get:
     \[
     2r = 2\sqrt{3} \cdot \sqrt{2} = 2\sqrt{6}
     \]
   - Solving for $r$, we get:
     \[
     r = \sqrt{6}
     \]

4. **Finding the height of triangle $ABC$**:
   - Let $O$ be the center of the circle.
   - $OB$ is the radius of the circle, so $OB = \sqrt{6}$.
   - Let $E$ be the midpoint of $AC$. Since $AC$ is a side of the square, $OE$ is half the side length of the square:
     \[
     OE = \frac{s}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
     \]
   - The height of the triangle $ABC$ is $BE = OB - OE$:
     \[
     BE = \sqrt{6} - \sqrt{3}
     \]

5. **Finding the area of triangle $ABC$**:
   - The base of triangle $ABC$ is the side length of the square, $s = 2\sqrt{3}$.
   - The height of triangle $ABC$ is $BE = \sqrt{6} - \sqrt{3}$.
   - The area of triangle $ABC$ is:
     \[
     \text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{3} \times (\sqrt{6} - \sqrt{3})
     \]
   - Simplifying, we get:
     \[
     \text{Area of } \triangle ABC = \sqrt{3} \times (\sqrt{6} - \sqrt{3}) = \sqrt{18} - \sqrt{9} = 3\sqrt{2} - 3
     \]

6. **Finding the total area of pentagon $ABCDE$**:
   - The area of the square $ACDE$ is $12$.
   - The area of triangle $ABC$ is $3\sqrt{2} - 3$.
   - The total area of the pentagon $ABCDE$ is:
     \[
     \text{Area of } ABCDE = 12 + (3\sqrt{2} - 3) = 9 + 3\sqrt{2}
     \]

The final answer is $\boxed{9 + 3\sqrt{2}}$