2. Let $a, b$ be two distinct real numbers, and satisfy $a^{2}=4 a$ $+3, b^{2}=4 b+3$. Then $\frac{a^{2}}{b}+\frac{b^{2}}{a}=\ldots$.
2. From the given, $a, b$ are the two unequal real roots of the equation $x^{2}-4 x-3=0$, so $a+b=4, a b=-3$. Therefore,
$$
\begin{array}{c}
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{a^{3}+b^{3}}{a b} \\
=\frac{(a+b)\left[(a+b)^{2}-3 a b\right]}{a b}=-\frac{100}{3} .
\end{array}
$$