[Geometry]
Problem: Let $T=T N Y W R$. The graph of $y=x^{2}+2 x-T$ intersects the $x$-axis at points $A$ and $M$, which are diagonally opposite vertices of square $A R M L$. Compute $[A R M L]$.

Solution: Note that the $x$-coordinates of $A$ and $M$ correspond to the two roots $r_{1}, r_{2}$ of $x^{2}+2 x-T$. If $s$ is the side length of square $A R M L$, then $A M=s \sqrt{2}=\left|r_{1}-r_{2}\right|=\sqrt{\left(r_{1}-r_{2}\right)^{2}}=$ $\sqrt{\left(r_{1}+r_{2}\right)^{2}-4 r_{1} r_{2}}=\sqrt{(-2)^{2}-4(-T)}=2 \sqrt{1+T}$. Thus $[A R M L]=s^{2}=2(1+T)$. With $T=36,[A R M L]=\mathbf{7 4}$.

Final answer: 74