4^x-1/2^x+2=128

Solution for 4^x-1/2^x+2=128 equation:

4^x-1/2^x+2=128

We move all terms to the left:

4^x-1/2^x+2-(128)=0


Domain of the equation: 2^x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

4^x-1/2^x-126=0

We multiply all the terms by the denominator


4^x*2^x-126*2^x-1=0

Wy multiply elements

8x^2-252x-1=0

a = 8; b = -252; c = -1;
Δ = b2-4ac
Δ = -2522-4·8·(-1)
Δ = 63536
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:

$\sqrt{\Delta}=\sqrt{63536}=\sqrt{5776*11}=\sqrt{5776}*\sqrt{11}=76\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-76\sqrt{11}}{2*8}=\frac{252-76\sqrt{11}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+76\sqrt{11}}{2*8}=\frac{252+76\sqrt{11}}{16}
$