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

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

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

We move all terms to the left:

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


Domain of the equation: 4c+12-c+8)!=0

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

4c-c+8)!=-12

c∈R


We add all the numbers together, and all the variables

-(-1c+3/4c+20)+8=0

We get rid of parentheses

1c-3/4c-20+8=0

We multiply all the terms by the denominator


1c*4c-20*4c+8*4c-3=0

Wy multiply elements

4c^2-80c+32c-3=0

We add all the numbers together, and all the variables

4c^2-48c-3=0

a = 4; b = -48; c = -3;
Δ = b2-4ac
Δ = -482-4·4·(-3)
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-28\sqrt{3}}{2*4}=\frac{48-28\sqrt{3}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+28\sqrt{3}}{2*4}=\frac{48+28\sqrt{3}}{8}
$