6. If $x_{1} 、 x_{2}$ are the roots of the equation $x^{2}+x-3=0$, then the value of $x_{1}^{3}-4 x_{2}^{2}+19$ is ( ).
(A)6
(B) 4
(C) 2
(D) 0
6. (D).

From the given, we have $x_{1}^{2}+x_{1}-3=0$. Thus, $x_{1}^{3}=3 x_{1}-x_{1}^{2}$. Also, $x_{2}^{2}=3-x_{2}$, so,
$$
\begin{array}{l}
x_{1}^{3}-4 x_{2}^{2}+19=3 x_{1}-x_{1}^{2}-4\left(3-x_{2}\right)+19 \\
=-\left(x_{1}^{2}+x_{1}-3\right)+4\left(x_{1}+x_{2}\right)+4=0
\end{array}
$$