How do you solve the equation $\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0$? $$
\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0
$$
My work:
$t = 8^x$
$\frac{1}{4}(t^{2+1})+5t - 3 = 0$
Then I don’t know how to do it, but I think I did it wrong.
 A: $\frac{1}{4}(t^2+1) +5t - 3 = 0$
$(t^2+1) + 20t - 12= 0$
$ t^2 + 20t - 11= 0$
Find it's roots using suitable method.
Here are some methods
And for this question you have to use quadratic formula.
Edit :
You have two different equation in your question. According to other solution is -
$\frac{1}{4}(8^{2x}\cdot8) +5\cdot8^x - 3 = 0$
$2\cdot8^{2x} +5\cdot8^x - 3 = 0$ 
Put $8^x = t$
$2t^2 + 5t - 3 = 0$
$2t^2 + 6t - 1t - 3 = 0$
$2t(t + 3) - 1(t + 3) = 0$
$(t + 3) (2t - 1) = 0$
$t = -3$ and $ t = \frac{1}{2}$ 
So $8^x = \frac{1}{2}$
$2^{3x} = 2^{-1}$
On equating powers,
$3x = -1$
$x = \frac{-1}{3}$
A: The title right now does not match the text, but presumably the equation to solve is
$${1\over4}8^{2x+1}+5\cdot8^x-3=0$$
The key here is that, when you let $t=8^x$, then $8^{2x+1}=8\cdot8^{2x}=8t^2$, not $t^{2+1}$.  You get a quadratic,
$$2t^2+5t-3=(2t-1)(t+3)=0$$
the roots of which are $t=1/2$ and $t=-3$.  The equation $8^x=1/2$, which can be rewritten as $2^{3x}=2^{-1}$, has solution $x=-1/3$.  The equation $8^x=-3$ has no solution (in real numbers, at least).  So $x=-1/3$ is the only solution.
