3/(12c-4)=15c+15

Solution for 3/(12c-4)=15c+15 equation:

3/(12c-4)=15c+15

We move all terms to the left:

3/(12c-4)-(15c+15)=0


Domain of the equation: (12c-4)!=0

We move all terms containing c to the left, all other terms to the right

12c!=4

c!=4/12

c!=1/3

c∈R


We get rid of parentheses

3/(12c-4)-15c-15=0

We multiply all the terms by the denominator


-15c*(12c-4)-15*(12c-4)+3=0

We multiply parentheses

-180c^2+60c-180c+60+3=0

We add all the numbers together, and all the variables

-180c^2-120c+63=0

a = -180; b = -120; c = +63;
Δ = b2-4ac
Δ = -1202-4·(-180)·63
Δ = 59760
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{59760}=\sqrt{144*415}=\sqrt{144}*\sqrt{415}=12\sqrt{415}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-12\sqrt{415}}{2*-180}=\frac{120-12\sqrt{415}}{-360}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+12\sqrt{415}}{2*-180}=\frac{120+12\sqrt{415}}{-360}
$