Help With: $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ I have this problem, $2(2^x- 1)x^2 + (2^{x^2}-2)x = 2^{x+1} -2$, but am stuck on how to solve it. I need to find all off the real solutions, but I have really no idea how to start. I noticed that it is a quadratic (kind of) if you move all the terms to one side, with $2(2^x- 1)$ being a, $(2^{x^2}-2)$ being b, and $-2^{x+1}+2$ being c. 
I still want to find the answer by myself, so can anyone help point me in the right direction? 
Thank you.
$\DeclareMathOperator{\sign}{sign}$


  Lemma 1: $$\sign(2^z-1)=\sign(z)$$


Proof: $2^z-1$ is strictly increasing and $2^0-1=0$. 


  Lemma 2: $$(2^a-1)b+(2^b-1)a=0\iff ab=0$$


Proof: By lemma 1, $\sign\left((2^a-1)b+(2^b-1)a\right)=\sign(ab)$. 


  Claim: $$x=0,\pm 1$$


Proof: By algebraic manipulation, the original becomes:

\begin{align*}
2(2^x-1)x^2+(2^{x^2}-2)x&amp;=2^{x+1}-2\\
(2^x-1)x^2+(2^{x^2-1}-1)x&amp;=2^x-1\\
(2^x-1)(x^2-1)+(2^{x^2-1}-1)x&amp;=0
\end{align*}

Therefore by lemma 2, $x(x^2-1)=0$.
Here's the graph of $2(2^x -1)x^2+(2^{x^2}-2)x - \left( 2^{x+1} - 2\right)$



$$x = \{ -1, 0, 1\}$$

Clearly the $x 2^{x^2}$ term dominates for large $|x|$, and guarantees that there are no more solutions, i.e., the set stated is maximal.