(1c/3)-4=-2(-1/6c+2)

Solution for (1c/3)-4=-2(-1/6c+2) equation:

(1c/3)-4=-2(-1/6c+2)

We move all terms to the left:

(1c/3)-4-(-2(-1/6c+2))=0


Domain of the equation: 6c+2))!=0

c∈R


We add all the numbers together, and all the variables

(+1c/3)-(-2(-1/6c+2))-4=0

We get rid of parentheses

1c/3-(-2(-1/6c+2))-4=0

We calculate fractions


6c^2/18c+()/18c-4=0

We multiply all the terms by the denominator


6c^2-4*18c+()=0

We add all the numbers together, and all the variables

6c^2-4*18c=0

Wy multiply elements

6c^2-72c=0

a = 6; b = -72; c = 0;
Δ = b2-4ac
Δ = -722-4·6·0
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-72}{2*6}=\frac{0}{12}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+72}{2*6}=\frac{144}{12}
=12
$