One, (20 points) Given that the roots of the equation $x^{2}-6 x-4 n^{2}-32 n=0$ are integers. Find the integer value of $n$.

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I. Solving for $x=3 \pm \sqrt{4 n^{2}+32 n+9}$.
Since the roots of the equation are integers, $4 n^{2}+32 n+9$ must be a perfect square. Let $4 n^{2}+32 n+9=m^{2}, m>0$. Then we have $(2 n+8+m)(2 n+8-m)=55$.
Since $55=1 \times 55=5 \times 11$ $=(-1) \times(-55)=(-5) \times(-11)$, solving these gives $n=10, n=0, n=-18, n=-8$. Therefore, the integer values of $n$ are $-18, -8, 0, 10$.