2. Given the equation $(a+1) x^{2}+(|a+2|-\mid a-$ $10 \mid) x+a=5$ has two distinct real roots, then $a$ can be ( .
(A) 5
(B) 9
(C) 10
(D) 11
2. A

The original equation is $(a+1) x^{2}+(|a+2|-|a-10|) x+a-5=0$.
According to the problem, we have
$$
\Delta=(|a+2|-|a-10|)^{2}-4(a+1)(a-5)>0 \text {, }
$$

which simplifies to $6 a>a^{2}+|a+2| \cdot|a-10|$.
When $a$ takes the values $9,10,11$, none of them satisfy (*) equation;
When $a=5$, it satisfies (*) equation.
In fact, when $a=5$, $a+1=6$.
$$
|a+2|-|a-10|=7-5=2, a-5=0 \text {. }
$$

Therefore, the original equation is $6 x^{2}+2 x=0$, whose roots are $0,-\frac{1}{3}$.