2c(3c-4)=4c-4

Solution for 2c(3c-4)=4c-4 equation:

2c(3c-4)=4c-4

We move all terms to the left:

2c(3c-4)-(4c-4)=0

We multiply parentheses

6c^2-8c-(4c-4)=0

We get rid of parentheses

6c^2-8c-4c+4=0

We add all the numbers together, and all the variables

6c^2-12c+4=0

a = 6; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·6·4
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{-(-12)-4\sqrt{3}}{2*6}=\frac{12-4\sqrt{3}}{12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*6}=\frac{12+4\sqrt{3}}{12}
$