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

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

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

We move all terms to the left:

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


Domain of the equation: 5)-4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


2x+(-(2x+(+4*3)/5-4x)*3)+10x/5-4x)*3)=0


We calculate terms in parentheses: +(-(2x+(+4*3)/5-4x)*3), so:

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

We add all the numbers together, and all the variables

-(2x+12/5-4x)*3

We multiply parentheses

-6x+12x-12/5*3

We multiply all the terms by the denominator


-6x*5*3+12x*5*3-12

Wy multiply elements

-90x*3+180x*3-12

Wy multiply elements

-270x+540x-12

We add all the numbers together, and all the variables

270x-12

Back to the equation:

+(270x-12)


We get rid of parentheses

2x+270x+10x/5-4x)*3)-12=0

We multiply all the terms by the denominator


2x*5+270x*5+10x-4x*3)-12)*5=0

We add all the numbers together, and all the variables

10x+2x*5+270x*5-4x*3)-12)*5=0

Wy multiply elements

-60x^2+10x+10x+1350x=0

We add all the numbers together, and all the variables

-60x^2+1370x=0

a = -60; b = 1370; c = 0;
Δ = b2-4ac
Δ = 13702-4·(-60)·0
Δ = 1876900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1876900}=1370$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1370)-1370}{2*-60}=\frac{-2740}{-120}
=22+5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1370)+1370}{2*-60}=\frac{0}{-120}
=0
$