2/3x+23/3=3/2x+4

Solution for 2/3x+23/3=3/2x+4 equation:

2/3x+23/3=3/2x+4

We move all terms to the left:

2/3x+23/3-(3/2x+4)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



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

x∈R


We get rid of parentheses

2/3x-3/2x-4+23/3=0

We calculate fractions


4x/54x^2+(-81x)/54x^2+46x/54x^2-4=0

We multiply all the terms by the denominator


4x+(-81x)+46x-4*54x^2=0

We add all the numbers together, and all the variables

50x+(-81x)-4*54x^2=0

Wy multiply elements

-216x^2+50x+(-81x)=0

We get rid of parentheses

-216x^2+50x-81x=0

We add all the numbers together, and all the variables

-216x^2-31x=0

a = -216; b = -31; c = 0;
Δ = b2-4ac
Δ = -312-4·(-216)·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-31}{2*-216}=\frac{0}{-432}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+31}{2*-216}=\frac{62}{-432}
=-31/216
$