If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take? 
If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?

This is what I have done:
Let $y = 3^x$
$$9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$$  $$\implies9y^2 + (t^2 - 4t - 2)y + 1 > 0$$
For the LHS to be greater than zero, $b^2 - 4ac$ has to be $\lt 0$, since coefficient of $y^2$ is greater than $0$(which will give us an upward opening parabola).
$$(t^2 - 4t - 2)^2  - 36< 0$$
The answer that I get is different from the correct answer. The correct answer to this is $t  \in \mathbb{R} - \{2\}$. What did I do wrong?

Alternately, by positivity of $3^x$ and AM-GM:
$$(t^2-4t-2)3^x+3^{2x+2}+1\ge (t^2-4t-2)3^x+2\sqrt{3^{2x+2}} = (t^2-4t+4) \cdot 3^x$$
with equality possible when $x=-1$.  So we need $t^2-4t+4 = (t-2)^2 > 0$ which means $t \neq 2$.
