(3)/(8)c-2=(2)/(3)c-12

Solution for (3)/(8)c-2=(2)/(3)c-12 equation:

(3)/(8)c-2=(2)/(3)c-12

We move all terms to the left:

(3)/(8)c-2-((2)/(3)c-12)=0


Domain of the equation: 8c!=0

c!=0/8

c!=0

c∈R



Domain of the equation: 3c-12)!=0

c∈R


We get rid of parentheses

3/8c-2/3c+12-2=0

We calculate fractions


9c/24c^2+(-16c)/24c^2+12-2=0

We add all the numbers together, and all the variables

9c/24c^2+(-16c)/24c^2+10=0

We multiply all the terms by the denominator


9c+(-16c)+10*24c^2=0

Wy multiply elements

240c^2+9c+(-16c)=0

We get rid of parentheses

240c^2+9c-16c=0

We add all the numbers together, and all the variables

240c^2-7c=0

a = 240; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·240·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*240}=\frac{0}{480}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*240}=\frac{14}{480}
=7/240
$