2/3c-4+2=3/2c=3

Solution for 2/3c-4+2=3/2c=3 equation:

2/3c-4+2=3/2c=3

We move all terms to the left:

2/3c-4+2-(3/2c)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R



Domain of the equation: 2c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

2/3c-(+3/2c)-4+2=0

We add all the numbers together, and all the variables

2/3c-(+3/2c)-2=0

We get rid of parentheses

2/3c-3/2c-2=0

We calculate fractions


4c/6c^2+(-9c)/6c^2-2=0

We multiply all the terms by the denominator


4c+(-9c)-2*6c^2=0

Wy multiply elements

-12c^2+4c+(-9c)=0

We get rid of parentheses

-12c^2+4c-9c=0

We add all the numbers together, and all the variables

-12c^2-5c=0

a = -12; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-12)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-12}=\frac{0}{-24}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-12}=\frac{10}{-24}
=-5/12
$