Solving $4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$ for real $x$ I was given the following equation:
$$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6; x\in \mathbb{R}$$
Here is how I tried to solve it:
$$2^{2x+2\sqrt{x^2-2}+1}-5\cdot 2^{\sqrt{x^2-2}}=12$$
$$2^{\sqrt{x^2-2}}(2^{2x+1+\sqrt{x^2-2}}-5)=2^2\cdot 3$$
Where $$2^{\sqrt{x^2-2}}=2^2 \land 2^{2x+1+\sqrt{x^2-2}}-5=3$$
$$\Updownarrow$$
$$\sqrt{x^2-2}=2 \land 2x+1+\sqrt{x^2-2}=3.$$
As well as we must note that $$x\in \left( -\infty ; -\sqrt2 \right]\cup \left[ \sqrt{2};\infty\right)$$
However
$$2x+1+\sqrt{x^2-2}=3 \Leftrightarrow -3x^2+8x-6=0$$
$$x_{1,2}=\frac{8\pm\sqrt{-8}}{6}$$
Thus, it's not satisfying given domain.
So I tried to use substitution
Let $\sqrt{x^2-2}=t, t\geq 0,$ then $x^2=t^2+2 \Leftrightarrow x=\pm \sqrt{t^2+2}$
If $$x=\sqrt{t^2+2}$$ then $$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$$
$$\Updownarrow$$ $$2^{2\sqrt{t^2+2}+2t}-5\cdot 2^{t-1}=6$$
$$\Updownarrow$$
$$2^{2\sqrt{t^2+2}+2\sqrt{t^2}}-5\cdot 2^{\sqrt{t^2}-1}=6$$
From here if I continued using substitution to make it into a quadratic equation, I would reach a result, however, I doubt this is the right way as I only had 10 minutes to solve this in class and I'm only up to high school. Can someone provide me with any solution?
 A: Comment: As you modified we have:
$2^{2x+1+2\sqrt{x^2-2}}-5\cdot 2^{\sqrt {x^2-2}}-12=0$
$2^{2x+1+2\sqrt{x^2-2}}=2^{2x+1}\cdot 2^{2\sqrt {x^2-2}}=2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2$
$2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2-5\cdot 2^{\sqrt{x^2-2}}-12=0$
Let $2^{\sqrt{x^2-2}}=y$ , we can write:
$(2^{2x+1})\cdot y^2-5y-12=0$
$y=\frac {5\pm\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$
Now you have to search whether following system of equations is  consistent or not:
$\begin {cases}\Delta=25+48\cdot 2^{2x+1}=k^2\\y=2^{\sqrt{x^2-2}}\end {cases}$
For solutions in $ R$ we take only positive solutions for y:
$y=\frac {5+\sqrt{25+48\cdot 2^{2x+1}}}{2\times 2^{2x+1}}$
Update:   we can see that:
$2^{2x+1}=11\rightarrow 2x+1\approx 3.5\rightarrow x=1.25$
$y=\frac{5+17}{2\times 11}=1=2^{\sqrt{x^2-2}}\Rightarrow X^2-2=0$
which gives $x=\sqrt 2\approx 1.4...$
1- $\sqrt 2$ gives $LHS=5.6$
2-$x=2$, gives $LHS=39.3$
3- In fact we must have $\sqrt 2<x< 2$, $x=1.6$ works:
$LHS\approx 12$
No need a complicated analytic solution; these three steps can be done using a calculator in less that ten minutes. What is more important is to find the domain.
