Example 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 $n$.

---

The translation maintains the original text's line breaks and format as requested.
Solution: Completing the square for $x$, we get
$$
(x-3)^{2}-\left(4 n^{2}+32 n+9\right)=0 \text {. }
$$

Then completing the square for $n$ (both steps of completing the square can be done at once), we get
$$
(x-3)^{2}-(2 n+8)^{2}+55=0 \text {. }
$$

Rearranging and factoring, we get
$$
(2 n+8+|x-3|)(2 n+8-|x-3|)=55 \text {. }
$$

Since $x$ and $n$ are both integers, and
$$
\begin{array}{l}
55=1 \times 55=5 \times 11 \\
=(-1) \times(-55)=(-5) \times(-11),
\end{array}
$$

Thus, we can obtain

Therefore, the integer values of $n$ are $-18, -8, 0, 10$ when the roots of the equation are both integers.

Note 1: Generally, the discriminant being a perfect square is only a necessary condition for the equation to have integer roots (in this problem, it is both necessary and sufficient). Here, we simultaneously find the roots of the equation, which is a general approach.
Note 2: The geometric background of this problem is to find the integer points on the hyperbola
$$
-\frac{(x-3)^{2}}{55}+\frac{(y+4)^{2}}{\frac{55}{4}}=1
$$

which has exactly 8 integer points.