It looks like we're told that
[tex]T(-2x^2)=3x^2+4x[/tex]
[tex]T\left(\dfrac12x+3\right)=-3x^2+3x-4[/tex]
[tex]T(2x^2-1)=-3x+4[/tex]
We use the fact that [tex]T[/tex] is linear to find [tex]T(1),T(x),T(x^2)[/tex]. First, we notice that
[tex]T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)[/tex]
We also have
[tex]T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)[/tex]
[tex]T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)[/tex]
So once we find [tex]T(1)[/tex], we can determine [tex]T(x)[/tex] and [tex]T(x^2)[/tex]. We have
[tex]T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4[/tex]
and using this we find
[tex]T(x)=12x^2+12x+16[/tex]
[tex]T(x^2)=-\dfrac32x^2-2x[/tex]
Then
[tex]T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)[/tex]
[tex]T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c[/tex]
