I wish it were a quadratic

We can solve this with some inspired addition. Our goal should be to create squared expressions on both sides that we can manipulate. Because the degree of $x$ on the LHS is $4,$ we will be adding multiples of $x^2$ and integers. Thus, $80x$ on the RHS will not change.

Since the $80x$ on the RHS cannot change, the constant on the RHS must be divisible by $25.$ Let's see what happens when we manipulate the constant on the RHS to be $100.$ $x^4=80x+36$ $x^4+64=80x+100$ This looks like it will work because the constant term on both sides of the equation is the square of an integer. We can now complete the square on the LHS. $x^4+64=80x+100$ $x^4+16x^2+64=16x^2+80x+100$ $(x^2+8)^2=4(4x^2+20x+25)$ The RHS also factors. $(x^2+8)^2=4(4x^2+20x+25)$ $(x^2+8)^2=(2(2x+5))^2$ We can now algebraically manipulate the equation to get a difference of squares and find the full factorization of the polynomial. $(x^2+8)^2=(2(2x+5))^2$ $(x^2+8)^2-(2(2x+5))^2=0$ $((x^2+8)+2(2x+5))((x^2+8)-2(2x+5))=0$ $(x^2+4x+18)(x^2-4x-2)=0$ Observing the two trinomial factors leads us to see that only the second factor has real roots. By Vieta's formula, these roots have a sum of $\boxed{4}.$