Surely this is trivial? $f(x)$ is a real polynomial, so $g(x)=f(x)+f(7-x)$ is a real polynomial, and in particular continuous, so by the intermediate value theorem (since $g(5)>7>g(4)$) there exists $u\in(4,5)$ such that $g(u)=7$.

Oh ,now I find a new solution: I think this solution has a little different to Diarmuid's solution We must prove the equality of $f(x)+f(7-x)=7$ has a root in real number. Suppose that $g(x)=f(x)=f(7-x)-7$ we know that $g(x)$ is a polynomial which its degree at most $2$ , $g(2)<0$ and $g(3)>0$ so that the equation $g(x)=0$, then it has at least one root. Excuse me for my bad question