<s>[INST] Functional equation: $ f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$ 
$f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17
\forall x \in \mathbb R $, then find the function $f(x)$

I have:
$f(15/4)= 16/3$ (both quadratics intersect at $x=1/2$)
$f(3)= 3$ (by substituting $x=0$ and $x=1$ and then solving the simultaneous equations obtained)
$f(5)= 7$
I am not getting anything fruitful from these.
Could anyone provide me a hint on how to proceed?
Edit: 
Someone had commented (now it's deleted) that we can assume $f(x) = ax +b$, how can we do that? I got the right answer using that.  Is it always okay to assume that way? When is it a reliable assumption?
 [/INST] Replace $x$ by $1-x$,
\begin{align*}
f((1-x)^2+(1-x)+3)+2f((1-x)^2−3(1-x)+5)&=6(1-x)^2−10(1-x)+17\\
f(x^2-3x+5)+2f(x^2+x+3)&=6x^2-2x+13\\
\end{align*}
Solving with $f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$, we have $f(x^2+x+3)=2x^2+2x+3$.
$f(x)=2x-3$.
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