10.2. Find all quadratic trinomials $f(x)$ for which the equality $f(2 x+1)=4 x^{2}+14 x+7$ holds.
Answer: $f(x)=x^{2}+5 x+1$.

First solution. Let $t=2 x+1$, then $x=1 / 2(t-1)$. Therefore, $f(t)=(t-1)^{2}+7(t-1)+7=t^{2}+5 t+1$.

Second: $4 x^{2}+14 x+7=\left(4 x^{2}+4 x+1\right)+10 x+5+1=(2 x+1)^{2}+5(2 x+1)+1$.

Third: quadratic trinomials are equal if their corresponding coefficients are equal. Denote $f(x)=a x^{2}+b x+c$, from the condition we get the relation $a(2 x+1)^{2}+b(2 x+1)+c=4 x^{2}+14 x+7$ from which the answer follows.

Remark. If the work contains the answer, the answer is verified, but the uniqueness of the answer is not noted, no more than 2 points.