2^2+2^x*2^1=128

Solution for 2^2+2^x*2^1=128 equation:

2^2+2^x*2^1=128

We move all terms to the left:

2^2+2^x*2^1-(128)=0

determiningTheFunctionDomain
2^x*2^1-128+2^2=0

We add all the numbers together, and all the variables

2^x*2^1-124=0

Wy multiply elements

4x^2-124=0

a = 4; b = 0; c = -124;
Δ = b2-4ac
Δ = 02-4·4·(-124)
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{31}}{2*4}=\frac{0-8\sqrt{31}}{8}
=-\frac{8\sqrt{31}}{8}
=-\sqrt{31}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{31}}{2*4}=\frac{0+8\sqrt{31}}{8}
=\frac{8\sqrt{31}}{8}
=\sqrt{31}
$