Because the problem only asks about the real roots of the polynomial, we can't apply Vieta's formulas directly. Instead, we recognize the coefficients from the expansion of $ (x-1)^5 $: \[(x-1)^5 = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1.\]Seeing this, we subtract $ x^5 $ from both sides, giving \[\begin{aligned} -x^5 + 5x^4 - 10x^3 + 10x^2 - 5x - 11 &= -x^5 \\ -(x-1)^5 - 12 &= -x^5 \\ (x-1)^5 + 12 &= x^5.\end{aligned}\]Hence, \[x^5 + (1 - x)^5 = 12.\]Let $ x = \frac{1}{2} + y $. Then $ 1 - x = \frac{1}{2} - y, $ so \[\left( \frac{1}{2} + y \right)^5 + \left( \frac{1}{2} - y \right)^5 = 12.\]This expands to \[5y^4 + \frac{5}{2} y^2 + \frac{1}{16} = 12.\]Consider the function \[f(y) = 5y^4 + \frac{5}{2} y^2 + \frac{1}{16}.\]Then $ f(0) = \frac{1}{16}, $ and $ f(y) $ is increasing on $ [0,\infty), $ so there is exactly one positive value of $ y $ for which $ f(y) = 12 $. Also, if $ f(y) = 12, $ then $ f(-y) = 12 $. This means that there are exactly two solutions in $ x, $ and if $ x $ is one solution, then the other solution is $ 1 - x $. Therefore, the sum of the solutions is $ \boxed{1} $.