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

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

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

We move all terms to the left:

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


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

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

4c-c)!=-12

c∈R


We add all the numbers together, and all the variables

-(-1c+3/4c+12)+8=0

We get rid of parentheses

1c-3/4c-12+8=0

We multiply all the terms by the denominator


1c*4c-12*4c+8*4c-3=0

Wy multiply elements

4c^2-48c+32c-3=0

We add all the numbers together, and all the variables

4c^2-16c-3=0

a = 4; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·4·(-3)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{19}}{2*4}=\frac{16-4\sqrt{19}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*4}=\frac{16+4\sqrt{19}}{8}
$