How to solve this?f=(2x^2-x-1)^10Calculate x_1+x_2+x_3+...+x_20.

$5$
$2 {x}^{2} - x - 1 = 2 \left(x + \frac{1}{2}\right) \left(x - 1\right)$ and
${\left(2 {x}^{2} - x - 1\right)}^{10} = {2}^{10} {\left(x + \frac{1}{2}\right)}^{10} {\left(x - 1\right)}^{10}$ so the sum of their $20$ roots is ${x}_{1} + {x}_{2} + \cdots + {x}_{20} = 10 \times 1 - 10 \times \frac{1}{2} = 5$