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

Solution for 2x/4x-1+(3/2x+1)-2/2x-1=0 equation:

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


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 2x+1)!=0

x∈R



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

2x/4x+3/2x-2/2x+1-2=0

We calculate fractions


4x^2/8x^2+(-8x+3)/8x^2+1-2=0

We add all the numbers together, and all the variables

4x^2/8x^2+(-8x+3)/8x^2-1=0

We multiply all the terms by the denominator


4x^2+(-8x+3)-1*8x^2=0

Wy multiply elements

4x^2-8x^2+(-8x+3)=0

We get rid of parentheses

4x^2-8x^2-8x+3=0

We add all the numbers together, and all the variables

-4x^2-8x+3=0

a = -4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·(-4)·3
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{7}}{2*-4}=\frac{8-4\sqrt{7}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*-4}=\frac{8+4\sqrt{7}}{-8}
$