2^x-4+(3/2^x)=0

Solution for 2^x-4+(3/2^x)=0 equation:

2^x-4+(3/2^x)=0


Domain of the equation: 2^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

2^x+3/2^x-4=0

We multiply all the terms by the denominator


2^x*2^x-4*2^x+3=0

Wy multiply elements

4x^2-8x+3=0

a = 4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·4·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*4}=\frac{4}{8}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8}
=1+1/2
$