4*2^2x-5*2^x+1=0

Solution for 4*2^2x-5*2^x+1=0 equation:

4*2^2x-5*2^x+1=0

Wy multiply elements

8x^2-10x+1=0

a = 8; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·8·1
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{17}}{2*8}=\frac{10-2\sqrt{17}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{17}}{2*8}=\frac{10+2\sqrt{17}}{16}
$