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{align*} -A^2 + B + C &= 0, \\ -AB + AC &= -80, \\ BC &= -36.\end{align*}From the second equation, $ B - C = \frac{80}{A} $. From the first equation, $ B + C = A^2 $. Squaring these equations, we get \begin{align*} B^2 + 2BC + C^2 &= A^4, \\ B^2 - 2BC + C^2 &= \frac{6400}{A^2}.\end{align*}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} $.