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

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

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

We move all terms to the left:

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


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

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

4c-c+4)!=-12

c∈R


We add all the numbers together, and all the variables

-(-1c+3/4c+16)+8=0

We get rid of parentheses

1c-3/4c-16+8=0

We multiply all the terms by the denominator


1c*4c-16*4c+8*4c-3=0

Wy multiply elements

4c^2-64c+32c-3=0

We add all the numbers together, and all the variables

4c^2-32c-3=0

a = 4; b = -32; c = -3;
Δ = b2-4ac
Δ = -322-4·4·(-3)
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{67}}{2*4}=\frac{32-4\sqrt{67}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{67}}{2*4}=\frac{32+4\sqrt{67}}{8}
$