2*2^x+1/4*4^x-320=0

Solution for 2*2^x+1/4*4^x-320=0 equation:

2*2^x+1/4*4^x-320=0


Domain of the equation: 4*4^x!=0

x!=0/1

x!=0

x∈R


Wy multiply elements

4x+1/4*4^x-320=0

We multiply all the terms by the denominator


4x*4*4^x-320*4*4^x+1=0

Wy multiply elements

64x^2*4-5120x*4+1=0

Wy multiply elements

256x^2-20480x+1=0

a = 256; b = -20480; c = +1;
Δ = b2-4ac
Δ = -204802-4·256·1
Δ = 419429376
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:

$\sqrt{\Delta}=\sqrt{419429376}=\sqrt{9216*45511}=\sqrt{9216}*\sqrt{45511}=96\sqrt{45511}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20480)-96\sqrt{45511}}{2*256}=\frac{20480-96\sqrt{45511}}{512}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20480)+96\sqrt{45511}}{2*256}=\frac{20480+96\sqrt{45511}}{512}
$