1/2c-4=2/3c-2

Solution for 1/2c-4=2/3c-2 equation:

1/2c-4=2/3c-2

We move all terms to the left:

1/2c-4-(2/3c-2)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R



Domain of the equation: 3c-2)!=0

c∈R


We get rid of parentheses

1/2c-2/3c+2-4=0

We calculate fractions


3c/6c^2+(-4c)/6c^2+2-4=0

We add all the numbers together, and all the variables

3c/6c^2+(-4c)/6c^2-2=0

We multiply all the terms by the denominator


3c+(-4c)-2*6c^2=0

Wy multiply elements

-12c^2+3c+(-4c)=0

We get rid of parentheses

-12c^2+3c-4c=0

We add all the numbers together, and all the variables

-12c^2-1c=0

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

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