Find $h(85)$ if $h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$ let : $h: \mathbb{R}\to \mathbb{R}$ ane for any real number 
$$h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$$
then :
$$h(85)=?$$
My Try: $$x=0:h(3)+2h(5)=17\\x=1:h(5)+2h(3)=13\\+\\3h(3)+3h(5)=30\\h(3)+h(5)=10$$
now ?
 A: Let $h(x) \equiv Ax+B$, then $$A(x^2+x+3)+B+2A(x^2-3x+5)+2B\equiv 6x^2-10x+17$$
$$
\left \{
\begin{align*}
  3A &= 6 \\
  -5A &= -10 \\
  13A+2B &= 17 \\
\end{align*}
\right.$$
On solving, $(A,B)=\left( 2, -\dfrac{9}{2} \right)$

Alternatively,


*

*$x=\dfrac{-1 \pm \sqrt{329}}{2} \implies x^2+x+3=85 \quad \text{and} \quad x^2-3x+5=89 \mp \sqrt{329}$

*$x=\dfrac{3 \pm \sqrt{329}}{2} \implies x^2-3x+5=85 \quad \text{and} \quad x^2+x+3=89 \pm \sqrt{329}$

*If you substitute $x=\dfrac{-1+\sqrt{329}}{2}$ and $x=\dfrac{3-\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89-\sqrt{329})$.

*If you substitute $x=\dfrac{-1-\sqrt{329}}{2}$ and $x=\dfrac{3+\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89+\sqrt{329})$.
A: Denote: $x^2+x+3=t$. Then:
$$h(t)+2h(t-4x+2)=6t-16x-1.$$
Plug $x=\frac{1}{2}$ to get:
$$h(t)+2h(t)=6t-9 \Rightarrow h(t)=2t-3.$$
Hence:
$$h(85)=2\cdot 85-3=167.$$
A: For a general way of solving this, look at my answer to this question.
When supposing it's a linear function, you also have to prove it can't be anything else, so your solution is complete.
