<s>[INST] 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.
 [/INST] $\frac{1}{4}(t^2+1) +5t - 3 = 0$
$(t^2+1) + 20t - 12= 0$
$ 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,
$3x = -1$
$x = \frac{-1}{3}$
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