jmc

We seek to factor $x^4-80x-36.$ Using the Integer Root theorem, we can determine that there are no integer roots, so we look for a factorization into two quadratics. Assume a factorization of the form $$(x^2+Ax+B)(x^2-Ax+C)=x^4-80x-36.$$(We take $A$ as the coefficient of $x$ in the first quadratic; then the coefficient of $x$ in the second quadratic must be $-A,$ to make the coefficent of $x^3$ in their product to be 0.)

Expanding, we get $$(x^2+Ax+B)(x^2-Ax+C)=x^4+(-A^2+B+C)x^2+(-AB+AC)x+BC.$$Matching coefficients, we get $$\begin{array}{rl}-A^2+B+C& =0,\\ -AB+AC& =-80,\\ BC& =-36.\end{array}$$From the second equation, $B-C=\frac{80}{A}.$ From the first equation, $B+C=A^2.$ Squaring these equations, we get $$\begin{array}{rl}{B}^2+2BC+C^2& =A^4,\\ B^2-2BC+C^2& =\frac{6400}{A^2}.\end{array}$$Subtracting these, we get $$A^4-\frac{6400}{A^2}=4BC=-144.$$Then $A^6-6400=-144A^2,$ so $A^6+144A^2-6400=0.$ This factors as $(A^2-16)(A^4+16A^2+400)=0,$ so $A=\pm 4.$

Taking $A=4,$ we get $B-C=20$ and $B+C=16,$ so $B=18$ and $C=-2$. Thus, $$x^4-80x-36=(x^2+4x+18)(x^2-4x-2).$$The quadratic factor $x^4+4x+18$ has no real roots. The quadratic factor $x^2-4x-2$ has real roots, and their sum is $\boxed{4}.$