In the following diagram, a semicircle is folded along a chord $AN$ and intersects its diameter $MN$ at $B$. Given that $MB : BN = 2 : 3$ and $MN = 10$. If $AN = x$, find $x^2$.
[asy]
size(120); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(pair P) {
dot(P,linewidth(3)); return P;
}
real r = sqrt(80)/5;
pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0];
draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2)));
label("$A$",D2(A),NW);
label("$B$",D2(B),SW);
label("$M$",D2(M),S);
label("$N$",D2(N),SE);
[/asy]
1. **Reflecting Point \( B \):**
   Reflect \( B \) over \( AN \) to obtain \( B' \). Since \( B' \) is the reflection of \( B \) over \( AN \), \( B' \) lies on the semicircle, and \( \angle ANB = \angle ANB' \). This implies that \( MA = AB' = AB \).

2. **Setting Up the Problem:**
   Given \( MB : BN = 2 : 3 \) and \( MN = 10 \), we can find the lengths of \( MB \) and \( BN \):
   \[
   MB = \frac{2}{5} \times 10 = 4, \quad BN = \frac{3}{5} \times 10 = 6
   \]

3. **Using the Pythagorean Theorem:**
   Let the altitude from \( A \) to \( MN \) be \( h \). Since \( MA \) is the radius of the semicircle, we have:
   \[
   MA^2 + AN^2 = (2^2 + h^2) + (h^2 + 8^2) = 10^2
   \]
   Simplifying the equation:
   \[
   2^2 + h^2 + h^2 + 8^2 = 10^2
   \]
   \[
   4 + h^2 + h^2 + 64 = 100
   \]
   \[
   2h^2 + 68 = 100
   \]
   \[
   2h^2 = 32
   \]
   \[
   h^2 = 16
   \]
   \[
   h = 4
   \]

4. **Finding \( MA \):**
   Using the Pythagorean theorem again for \( MA \):
   \[
   MA^2 = 2^2 + 4^2
   \]
   \[
   MA^2 = 4 + 16
   \]
   \[
   MA^2 = 20
   \]

5. **Finding \( AN \):**
   Since \( AN \) is the hypotenuse of the right triangle with legs \( MA \) and \( h \):
   \[
   AN^2 = MA^2 + h^2
   \]
   \[
   AN^2 = 20 + 16
   \]
   \[
   AN^2 = 36
   \]

The final answer is \( \boxed{36} \).