3x+2=((5x+4)/4)+x-2

Solution for 3x+2=((5x+4)/4)+x-2 equation:

3x+2=((5x+4)/4)+x-2

We move all terms to the left:

3x+2-(((5x+4)/4)+x-2)=0


Domain of the equation: 4)+x-2)!=0

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


3x*4)+x-2)-(((5x+4)+2*4)+x-2)=0

Wy multiply elements

12x^2+8x=0

a = 12; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·12·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*12}=\frac{-16}{24}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*12}=\frac{0}{24}
=0
$