3. Let $x_{1}, x_{2}$ be the two real roots of the equation $x^{2}+x-4=0$. Then $x_{1}^{3}-5 x_{2}^{2}+10=(\quad)$.
(A) -29
(B) -19
(C) -15
(D) -9
3. B.

Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation $x^{2}+x-4=0$, therefore,
$$
\begin{array}{l}
x_{1}^{3}=x_{1} x_{1}^{2}=x_{1}\left(4-x_{1}\right)=4 x_{1}-x_{1}^{2} \\
=4 x_{1}+x_{1}-4=5 x_{1}-4, \\
-5 x_{2}^{2}=-5\left(4-x_{2}\right)=-20+5 x_{2} .
\end{array}
$$

Then $x_{1}^{3}-5 x_{2}^{2}+10=5\left(x_{1}+x_{2}\right)+10-24$
$$
=-5+10-24=-19 \text {. }
$$