8=3/4c+12-c+

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

8=3/4c+12-c+

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)+8=0

We get rid of parentheses

1c-3/4c+8=0

We multiply all the terms by the denominator


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

Wy multiply elements

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}
$