Values of $a$ that satisfy inequality 
Find the values of $a$ for which $9^{x+1}+(a^2-4a-2)\cdot3^x+1>0,$ for all $x\in \mathbb R$.

This is how I did it:
Since $9^{x+1}+1 > 0\text{ for all } x\in \mathbb R, (a^2-4a-2)3^x \text{ should be greater than } 0$, so
$$(a^2-4a-2)3^x>0\implies (a^2-4a-2)>0$$
$$\implies a\in[-\infty,2-\sqrt6]\cup[2+\sqrt6,\infty]$$
However, the correct answer is $a\in\mathbb R-\{2\}$.
What did I do wrong?
P.S. Although I have asked this question before, but now I want to know what's wrong in my new solution. So I don't need some other solution.

Hint
Suppose you rewrite $$9^{x+1}+(a^2-4a-2)\cdot3^x+1$$ as $$9 \times 9^x+(a^2-4a-2)\cdot3^x+1=9 \times 3^{2x}+(a^2-4a-2)\cdot3^x+1$$ and now define $y=3^x$. So the expression becomes $$9y^2+(a^2-4a-2)y+1$$ and you want it to be positive.
I am sure that you can take from here.
