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-4)=0


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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

4c^2-3=0

a = 4; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·4·(-3)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*4}=\frac{0-4\sqrt{3}}{8}
=-\frac{4\sqrt{3}}{8}
=-\frac{\sqrt{3}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*4}=\frac{0+4\sqrt{3}}{8}
=\frac{4\sqrt{3}}{8}
=\frac{\sqrt{3}}{2}
$