${\text{f:R→R, f(x}}^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17 \forall x\in R then the value of f(100) is$

$$\begin{gathered} \text{obviously f is a linear polynomial} \\ {\text{let f(x)=ax+b hence f(x}}^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17 \end{gathered}$$

$$\begin{gathered} \left[a(x^2+x+3)+b \right]+2\left[a(x^2-3x+5)+b\right]=6x^2-10x+17 \\ comparing the coefficient of x^2 and constants \end{gathered}$$

$\Rightarrow a+2a=6 and 3a+b+10a+2b=17$

$$\begin{gathered} \Rightarrow 3a=6 and 13a+3b=17 \\ \Rightarrow a=2 and 26+3b=17 \\ b=-3 \\ \therefore f\left(x\right)=2x-3 \\ \Rightarrow f\left(100\right)=197 \end{gathered}$$