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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 $$\begin{array}{rl}-x^5+5x^4-10x^3+10x^2-5x-11& =-x^5\\ -(x-1{)}^5-12& =-x^5\\ (x-1{)}^5+12& =x^5.\end{array}$$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 $\boxed{1}.$