3/412c-4=15c+15

Solution for 3/412c-4=15c+15 equation:

3/412c-4=15c+15

We move all terms to the left:

3/412c-4-(15c+15)=0


Domain of the equation: 412c!=0

c!=0/412

c!=0

c∈R


We get rid of parentheses

3/412c-15c-15-4=0

We multiply all the terms by the denominator


-15c*412c-15*412c-4*412c+3=0

Wy multiply elements

-6180c^2-6180c-1648c+3=0

We add all the numbers together, and all the variables

-6180c^2-7828c+3=0

a = -6180; b = -7828; c = +3;
Δ = b2-4ac
Δ = -78282-4·(-6180)·3
Δ = 61351744
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:

$\sqrt{\Delta}=\sqrt{61351744}=\sqrt{64*958621}=\sqrt{64}*\sqrt{958621}=8\sqrt{958621}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7828)-8\sqrt{958621}}{2*-6180}=\frac{7828-8\sqrt{958621}}{-12360}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7828)+8\sqrt{958621}}{2*-6180}=\frac{7828+8\sqrt{958621}}{-12360}
$