<s>[INST] 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?
 [/INST] Given  $$9^{x+1}+(t^2-4t-2)3^x+1>0$$
We can write it as $$3^x\bigg[9\cdot 3^x+(t^2-4t-2)+\frac{1}{3^x}\bigg]>0$$
So $$\underbrace{3^x}_{>0} \bigg[\bigg(3\cdot 3^{\frac{x}{2}}-\frac{1}{3^{\frac{x}{2}}}\bigg)^2+(t-2)^2\bigg]>0$$
So for Above statement is True $\forall \ x \; \mathbb{R},$ If $t\in \mathbb{R}-\{2\}$
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