$3(2x^2+x-2)^2=8x^2+4x-9$

Question

I should solve the following biquadratic equation with changing of variables (e.g., $x^2=z$). I do not see how to do it, and I think the answer in my book does not cover all solutions: $1$ and $-\dfrac{3}{2}$.

$3(2x^2+x-2)^2=8x^2+4x-9$

Answer 1

Partial answer. Observe $8x^2+4x-9=4(2x^2+x)-9.$ Now, we can let $y:=2x^2+x$ which gives $$3(y-2)^2=4y-9.$$ This then is equivalent to $$3y^2-12y+12=4y-9\Rightarrow3y^2-16y+21=0\Rightarrow(3y-7)(y-3)=0\Rightarrow y=7/3,\quad y=3.$$

Now we can undo the substitutions and we should get the desired answers.

Answer 2

Hint: Left minus right-hand side can factorized into $$(x-1) (2 x+3) \left(6 x^2+3 x-7\right)=0$$