Solutions of an equation Problem:

Find all real solutions for $x$ in
  $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$

I know that you can move all the terms to one side, then divide by $2$ and factor, but what next?

We seek all real solutions for $2(2^x-1)x^2+\left(2^{x^2}-2\right)x=2^{x+1}-2$.
Let's rewrite our equation into a very interesting form.
\begin{align}
2(2^x-1)x^2+\left(2^{x^2}-2\right)x&=2^{x+1}-2\\
2(2^x-1)x^2+\left(2^{x^2}-2\right)x-2^{x+1}+2&=0&&\text{Set equation = 0}\\
(2^x-1)x^2+\left(2^{x^2-1}-1\right)x-2^x+1&=0&&\text{Divide all terms by 2}\\
(2^x-1)x^2+\left(2^{x^2-1}-1\right)x-(2^x-1)&=0&&\text{Factor out $-1$}\\
(2^x-1)(x^2-1)+\left(2^{x^2-1}-1\right)x&=0&&\text{Factor out $2^x-1$}\\
(2^x-1)y+x(2^y-1)&=0\tag{1}\label{eq}&&\text{Replace $x^2-1$ with $y$}
\end{align}
Notice the symmetry of the left-hand side of Equation \ref{eq}. Also notice that if $x=0$, then $2^x-1=0$ as well.
The solutions of Equation \ref{eq} must be values such that $x=0$ or $y=x^2-1=0$. Therefore the solutions are $x=0, 1, -1$.
