5. The positive integer solutions of the equation $2 x^{2}+y^{2}-2 x y-4 x-30=0$ are $\qquad$
5. $(5,10),(7,4),(7,10)$.

Since $y^{2}-2 x y+\left(2 x^{2}-4 x-30\right)=0$ is a quadratic equation in $y$ with positive integer solutions, we have:
$$
34-(x-2)^{2}=k^{2}
$$

is a perfect square.
Thus, $k^{2}$ can only be $25,16,9,4,1,0$.
From $34-(x-2)^{2}=25 \Rightarrow(x-2)^{2}=9$, we get $x=5, y=10$;
From $34-(x-2)^{2}=9 \Rightarrow(x-2)^{2}=25$, we get $x=7, y=10$ or $x=7, y=4$;
When $k^{2}$ is $16,4,1,0$, there are no positive integer solutions.
Therefore, the positive integer solutions are $(5,10),(7,4),(7,10)$.