6. Let the parabola $y=x^{2}+x-1$ intersect the $x$-axis at points $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$. Then $10 x_{2}^{3}+4 x_{1}^{5}-58=(\quad)$.
(A) 0
(B) -100
(C) -2
(D) 82
6. B.

From the given conditions, we have
$$
\begin{array}{l}
x_{1}^{2}=1-x_{1} \\
\Rightarrow x_{1}^{5}=\left(1-x_{1}\right)^{2} x_{1}=\left(1-2 x_{1}+1-x_{1}\right) x_{1} \\
\quad=2 x_{1}-3 x_{1}^{2}=5 x_{1}-3 .
\end{array}
$$

Similarly, $x_{2}^{3}=2 x_{2}-1$.
Then $10 x_{2}^{3}+4 x_{1}^{5}-58$
$$
\begin{array}{l}
=10\left(2 x_{2}-1\right)+4\left(5 x_{1}-3\right)-58 \\
=20\left(x_{1}+x_{2}\right)-80=20 \times(-1)-80 \\
=-100
\end{array}
$$