6. Given that the roots of the equation $x^{2}-6 x-4 n^{2}-32 n=0$ are integers. Find the integer value of $n$.
(2004, National Junior High School Mathematics League)
(Tip: $\Delta=4\left(4 n^{2}+32 n+9\right)$. Since the roots of the equation are all integers, $4 n^{2}+32 n+9$ is a perfect square. Let $4 n^{2}+32 n+9=m^{2}$ (where $m$ is a natural number). Then
$$
\begin{array}{l}
(2 n+8+m)(2 n+8-m)=55 . \\
\text { Also } 55=( \pm 1) \times( \pm 55)=( \pm 5) \times( \pm 11), \\
2 n+8+m>2 n+8-m,
\end{array}
$$

Therefore, four systems of equations can be formed. Solving them yields $n=-18,-8$, $0,10$.