5. Given the function $f(x)$ satisfies for all real numbers $x$, $y$,
$$
f(x)+f(2 x+y)+5 x y=f(3 x-y)+2 x^{2}+1 \text {. }
$$

Then the value of $f(10)$ is ( ).
(A) -49
(B) -1
(C) 0
(D) 25
5. A.

Let $x=10, y=5$, we get $f(10)+f(25)+250=f(25)+200+1$. Therefore, $f(10)=-49$.