10. Given that $n$ is an integer, and the equation
$$
(n+1)^{2} x^{2}-5 n(n+1) x+\left(6 n^{2}-n-1\right)=0
$$
$(n \neq-1)$ has two integer roots. Then $n=$ $\qquad$ .
10.0 or -2.
Decompose the equation into
$$
[(n+1) x-(3 n+1)][(n+1) x-(2 n-1)]=0 \text {. }
$$

The roots are obtained as
$$
\begin{array}{l}
x_{1}=\frac{3 n+1}{n+1}=3-\frac{2}{n+1}, \\
x_{2}=\frac{2 n-1}{n+1}=2-\frac{3}{n+1} .
\end{array}
$$

Therefore, for $x_{1}$ to be an integer, $n+1$ can only be $\pm 1$, $\pm 2$; for $x_{2}$ to be an integer, $n+1$ can only be $\pm 1$, $\pm 3$.

Thus, for both $x_{1}$ and $x_{2}$ to be integers, $n+1$ can only be $\pm 1$, i.e., $n=0$ or -2.