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


I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
As per dxiv's hint: by elementary methods, the equation is equivalent to

$$f(x)\stackrel{\text{def}}{=}(2^x−1)(x^2-1)+(2^{x^2-1}−1)x=0$$
Note that, for any $y$, $2^y-1$ and $y$ have the same sign (positive, negative, or zero); write $\mathrm{sgn}\,y$ for the sign of $y$.
Then

$$\mathrm{sgn}[(2^x-1)(x^2-1)]=\mathrm{sgn}(2^x-1)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)$$
and likewise
$$\mathrm{sgn}[(2^{x^2-1}-1)x]=\mathrm{sgn}(x^2-1)\mathrm{sgn}(x)$$

Now, if $\mathrm{sgn}(y)=\mathrm{sgn}(z)$, then $\mathrm{sgn}(y+z)=\mathrm{sgn}(y)=\mathrm{sgn}(z)$. Consequently
$$\mathrm{sgn}[f(x)]=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x+1)\mathrm{sgn}(x-1)\text{.}$$
By the zero product property and the fact that $\mathrm{sgn}(y)=0$ iff $y=0$, we find that $f(x)=0$ iff $x\in\{-1,0,1\}$.
$\begin{align*} 2\left (2^{x}-1 \right )x^{2}+\left(2^{x^{2}}-2\right)x&amp;=2^{x}.2-2\\ 2\left(2^{x}-1\right)x^{2}-\left(2^{x^{2}}-2\right)x&amp;=2(2^{x}-1) \end{align*}$

Let $2^{x}-1=a$ and $2^{x^{2}}-2=b$ and equation becomes:
$2ax^{2} + bx=2a$ or $2ax^{2}+bx-2a=0$. It's a quadratic equation. The quadratic equation has real solution if and only if $b^{2}-4ac&gt;0$.