If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$. 
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.

Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$. 
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values. 
Thank you very much for helping.

Consider a linear function $ f(x)=ax+b$
$$ f(x^2+x) = ax^2+ax+b$$
$$ f(x^2-3x+2)= ax^2-3ax +2a+b$$
$$ f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$
$$a=3, b=-4$$
$$ f(x) = 3x-4$$
$$f(2016)=6044$$
