This is a rather weird shape that even my teacher couldn't solve. (Lets ignore the faulty touchscreen)
The red letters are the angles and A.C.E are perfectly straight (Colinear)
Is there an way to solve this question? The answer is 2 but we couldn't find it.
I am just an early high schooler.
Construct triangle $DEB$. The sides $DE=BE=10, DB=12$ (Pythagorean theorem). Let $O$ be point of intersection of $AE$ and $DB$. Then $EO$ is the height in isosceles $\triangle DEB$.
We also know that the diagonals of square bisect so $OB=OC=6$. Thus, $EO=\sqrt{10^2-6^2}=8$.
Finally, $x=EO-OC=2.$