(2x/3x)+(x/2)=0

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

(2x/3x)+(x/2)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2x/3x)+(+x/2)=0

We get rid of parentheses

2x/3x+x/2=0

We calculate fractions


3x^2/6x+4x/6x=0

We multiply all the terms by the denominator


3x^2+4x=0

a = 3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·3·0
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{-(4)-4}{2*3}=\frac{-8}{6}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*3}=\frac{0}{6}
=0
$