Some guys kept on summing!

$\begin{aligned} 5x^4-10x^3+10x^2-5x-11 & = 0 & \small \color{#3D99F6} \text{Let }x^5-12 \text{ subtracts both sides.} \\ x^5 - 5x^4+10x^3-10x^2+5x-1 & = x^5-12 \\ (x-1)^5 & = x^5 - 12 \\ \implies x^5 - (x-1)^5 - 12 & = 0 & \small \color{#3D99F6} \text{Let } u = x-\frac 12 \\ \left(u+\frac 12\right)^5 - \left(u-\frac 12\right)^5 - 12 & = 0 \\ 5u^4+\frac 52 u^2 + \frac 1{16} - 12 & = 0 \\ 80u^4 + 40u^2 - 191 & = 0 \end{aligned}$

$\begin{aligned} \implies u^2 & = \frac {-40+\sqrt{62720}}{160} = \frac {7\sqrt 5}{10} - \frac 14 & \small \color{#3D99F6} \text{Since }u^2 > 0 \\ \implies u & = \begin{cases} u_1= + \color{#3D99F6} \alpha & \implies x_1 = \dfrac 12 + \color{#3D99F6} \alpha \\ u_2 = - \color{#3D99F6} \alpha & \implies x_2 = \dfrac 12 - \color{#3D99F6} \alpha \end{cases} & \small \color{#3D99F6} \text{where }\alpha = \sqrt{\frac {7\sqrt 5}{10} - \frac 14} \end{aligned}$

Therefore, the sum of real roots $x_1 + x_2 = \boxed 1$ .

For $f(x)=x^4-2x^3+2x^2-x-\frac{11}{5}$ we have $f'(x)=4x^3-6x^2+4x-1,f''(x)=12x^2-12x+4$ and $f'''(x)=24x-12$ . Since $f'''(\frac{1}{2})=f'(\frac{1}{2})=0$ , the function $f(x)$ is even at $\frac{1}{2}$ . Since $f''(x)>0$ throughout, the graph of $f(x)$ is convex, and $f(x)$ has exactly two real roots $\frac{1}{2}\pm a$ . Thus the answer is $\boxed{1}$ .