13. Given a quadratic equation in $x$
$$
(6-k)(9-k) x^{2}-(117-15 k) x+54=0
$$

both roots of which are integers. Find all real values of $k$ that satisfy the condition.
13. The original equation can be transformed into
$$
[(6-k) x-9][(9-k) x-6]=0 \text {. }
$$

Since this equation is a quadratic equation in $x$, we have $-k \neq 6, k \neq 9$. Therefore, we have
$$
x_{1}=\frac{9}{6-k}, x_{2}=\frac{6}{9-k} \text {. }
$$

Eliminating $k$, we get $x_{1} x_{2}-2 x_{1}+3 x_{2}=0$, which simplifies to
$$
\left(x_{1}+3\right)\left(x_{2}-2\right)=-6 \text {. }
$$

Since $x_{1}$ and $x_{2}$ are both integers, we have
$$
\left\{\begin{array}{l}
x_{1}+3=-6,-3,-2,-1,1,2,3,6 \\
x_{2}-2=1,2,3,6,-6,-3,-2,-1 .
\end{array}\right.
$$

Thus, $x_{1}=-9,-6,-5,-4,-2,-1,0,3$.
Clearly, $x_{1} \neq 0$.
Also, $k=\frac{6 x_{1}-9}{x_{1}}=6-\frac{9}{x_{1}}$. Substituting $x_{1}=-9,-6,-5,-4,-2,-1,3$ respectively, we get
$$
k=7, \frac{15}{2}, \frac{39}{5}, \frac{33}{4}, \frac{21}{2}, 15,3 .
$$