Because the square of all primes other than 2 and 3 are congruent to one mod 24, we can equate $2x^2-x-36=1 ($mod $24)$, which simplifies to $x=13 ($mod $24)$. I'm not where to go from here

No need for modulos. $$2x^2-x-36 = (2x-9)(x+4) = p^2$$Then we must have $2x-9=x+4 \implies x = 13.$ Testing $13$ we see the expression equates to $17^2$ which works. We must also test for the case in which one factor is $1.$ Obviously $x=-3$ does not work but $x=5$ does work, so we have $x\in \{5, 13\}.$ @below fixed but i really forgot a case on a 1st grade problem

svyn wrote: No need for modulos. $$2x^2-x-36 = (2x-9)(x+4) = p^2$$Then we must have $2x-9=x+4 \implies x = 13.$ Testing $13$ we see the expression equates to $17^2$ which works. Any other value of $x$ does not work, for the expression is no longer the square of a prime. Our only answer is $13.$ Your solution is incomplete, even $(x,p) = (5,3)$ works

$(2x-9)(x+4)=p^2 \Rightarrow 2x-9=p^2,x+4=1$ or $2x-9=x+4=p$ or $2x-9=1,x+4=p^2$ or $2x-9=-p^2,x+4=-1$ or $2x-9=x+4=-p$ or $2x-9=-1,x+4=-p^2$. Working through all of them, we only get $(x,p)=(5,3)$ or $(13,17)$.