2^-x+1=-3/4x+3

Solution for 2^-x+1=-3/4x+3 equation:

2^-x+1=-3/4x+3

We move all terms to the left:

2^-x+1-(-3/4x+3)=0


Domain of the equation: 4x+3)!=0

x∈R


We add all the numbers together, and all the variables

-1x-(-3/4x+3)=0

We get rid of parentheses

-1x+3/4x-3=0

We multiply all the terms by the denominator


-1x*4x-3*4x+3=0

Wy multiply elements

-4x^2-12x+3=0

a = -4; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·(-4)·3
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{3}}{2*-4}=\frac{12-8\sqrt{3}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{3}}{2*-4}=\frac{12+8\sqrt{3}}{-8}
$