We are given this plot
from which I extract that $f(x)=\frac{x^2}{1600}-\frac{x}{4}+125$. I am asked to find the height at which the ground should be leveled, in such a way that that the exceeding ground from one side fills the lack of it in the other. Tying to find a procedure, I end up trying to solve the following equation, that is too complicated to be correct:
$$x (\frac{x^2}{1600}-\frac{x}{4}+125) \int_0^x f(x) dx = \int_0^{600}f(x) dx - (600-x)(\frac{x^2}{1600}-\frac{x}{4}+125)$$
This seems to be a nonsense, and any other approach I've tried leads nowhere. Sorry for bothering you.
Guide:
If the area under the graph is preserved, then we have
$$600 h = \int_0^{600} f(x) \, dx$$
Just evaluate the integral and then divide by $600$.