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?
 A: With the substitution $y= 3^x$, as you mention, we get the equivalent inequality:
$$9y^2+(t^2-4t-2)y>0$$
but with the added condition $y> 0$ as $3^x > 0$.
Hence we see that if $b=t^2-4t-2\ge 0$, the LHS is positive for all allowable $y$, so we need to worry only about $b< 0$.  Here using the discriminant condition, it is sufficient to show that 
$$b^2< 4ac \iff (t^2-4t-2)^2< 36 \iff -6 < t^2-4t - 2 < 6$$
As we need only consider $b< 0$, only the left inequality matters. So we have $t^2-4t+4 = (t-2)^2 > 0 \implies t \neq 2$.
A: 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 $t \neq 2$.
A: Y needs to be positive, and this will give you another inequality for t
A: 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 $t\in \mathbb{R}-\{2\}$
A: You need $b^2 - 4ac$ not $b - 4ac$.
