If 9^x+1+(a^2-4a-2) 3^x +1>0 for all x ∈ R, then

We have,

$9^{x+1}+(a^2-4a-2) 3^x +1>0$ for all x ∈ R

$⇒9y^2+(a^2-4a-2) y + 1 > 0$ for all $y > 0$, where $y = 3^x$

$⇒y(9y+\frac{1}{y}+a^2-4a-2)>0$ for all $y>0$

$⇒(3\sqrt{y}-\frac{1}{\sqrt{y}})^2+(a-2)^2>0$ for all $y>0$

$⇒a≠2⇒a ∈ R-\{2\}$