4-12+c+3/4c=8

Solution for 4-12+c+3/4c=8 equation:

4-12+c+3/4c=8

We move all terms to the left:

4-12+c+3/4c-(8)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R


We add all the numbers together, and all the variables

c+3/4c-16=0

We multiply all the terms by the denominator


c*4c-16*4c+3=0

Wy multiply elements

4c^2-64c+3=0

a = 4; b = -64; c = +3;
Δ = b2-4ac
Δ = -642-4·4·3
Δ = 4048
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:

$\sqrt{\Delta}=\sqrt{4048}=\sqrt{16*253}=\sqrt{16}*\sqrt{253}=4\sqrt{253}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{253}}{2*4}=\frac{64-4\sqrt{253}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{253}}{2*4}=\frac{64+4\sqrt{253}}{8}
$