$e^{6x}-e^{4x}-2e^{3x}-12e^{2x}+e^x+1=0$

${\left(e^{3x}\right)}^2-2e^{3x}+1-e^{4x}+e^x=12e^{2x}$

$\Rightarrow {\left(e^{3x}-1\right)}^2-e^x\left(e^{3x}-1\right)=12e^{2x}$

$\Rightarrow \left(e^{3x}-1\right)\left(e^{3x}-1-e^x\right)=12e^{2x}$

$\Rightarrow \frac{\left(e^{3x}-1-e^x\right)}{e^{2x}}=\frac{12}{\left(e^{3x}-1\right)}$

$\Rightarrow e^x-e^{-x}-e^{-2x}=\frac{12}{e^{3x}-1}$

Assume, $f\left(x\right)=e^x-e^{-x}-e^{-2x} \& g\left(x\right)=\frac{12}{e^{3x}-1}$

$f\left(x\right)=e^x-e^{-x}-e^{-2x}$

$f^{\text{'}}\left(x\right)=e^x+e^{-x}+2e^{-2x}$

Clearly, $f^{\text{'}}\left(x\right)>0\forall x\in R \left(\because e^x \& e^{-x}>0\forall x\in R\right)$

So, here $f\left(x\right)$ is increasing and $f\left(0\right)=-1$and clearly $g\left(x\right)$ is decreasing function because it is reciprocal function.

From the graph, we can say that number of real roots $=2$