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.

First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily,
When $x = \dfrac{-1 - \sqrt{8065}}{2}$:


*

*$x^2 + x = 2016$

*$x^2 - 3x + 2 = 2020 + 2\sqrt{8065} = a$ (say)

*$9x^2 - 15x = 18156 + 12\sqrt{8065}$


$$f(2016) + 2f(a) = 18156 + 12\sqrt{8065}$$
When $x = \dfrac{3 + \sqrt{8065}}{2}$:


*

*$x^2 + x = a$

*$x^2 - 3x + 2 = 2016$

*$9x^2 - 15x = 18144 + 6\sqrt{8065}$.


$$f(a) + 2f(2016) = 18144 + 6\sqrt{8065}$$
From the two equations,
$$4f(2016) - f(2016) = 2(18144) - 18156$$
$$\boxed{f(2016) = 6044}$$
