Because $x_{1,2}$ are roots of $x^2-x-1=0$, we find $x_1^2=x_1+1$ and $x_2^2=x_2+1$. Meanwhile, by Vieta's theorem, we find $x_1 + x_2=1$. Therefore \begin{align*} 2x_1^5 + 5x_2^3 &= 2x_1(x_1+1)^2 + 5x_2(x_2+1)\\ &= 2x_1(x_1^2 + 2x_1+1) + (5x_2^2 + 5x_2)\\ &= 2x_1((x_1+1) + 2x_1+1) + (5(x_2 + 1) + 5x_2)\\ &= 6x_1^2 + 4x_1 + 10x_2 + 5\\ &= 6(x_1+1) + 4x_1 + 10x_2 + 5\\ &= 10(x_1+x_2)+ 11\\ &= \boxed{21} \end{align*}