Don't use algebra

$\overline{AF}=\overline{FE} \ \text{and} \ \angle AFG =\angle GFE \implies \overline{FG} \perp \overline{A E}$ $\angle nGF =\angle FAE$ $\triangle GnF\cong\triangle ADE$ The length of $\overline{GF}=\overline{AE}=15,\overline{DE}=\sqrt{15^2-12^2}=9,\overline{CE}=12-9=3$

Let $\overline{GJ}\perp\overline{AD}$ . $\overline{GJ}=\overline{AD}=12$ $\angle GJF=\angle ADE=90^{\circ}$ $\angle DAE=\angle JGF=90^{\circ}-\angle AFG$ $\triangle GJF\cong\triangle ADE(ASA)$ So, $\overline{GF}=\overline{AE}=15,\overline{DE}=\sqrt{15^2-12^2}=9,\overline{CE}=12-9=3$

We know EF + DF must be 12 as they are one side length of the square. With Phythagoras we know that DE² = EF² - DF². In the right triangle DEF the angle DFE is (through folding) twice as big as the angle BGF - 90°, so angle DEF is 2arccos(12/15)=73.74°. Then we can say that DF = EF*cos(73.74°) = 0.28EF. Now we can solve the equations from the beginning for EF, DF and DE and get EF=9.375, DF=2.625 and DE=9. DE and CE together form one side length of the square so that CE = 12-DE = 3.