6=-2c(12-2c)+12

Solution for 6=-2c(12-2c)+12 equation:

6=-2c(12-2c)+12

We move all terms to the left:

6-(-2c(12-2c)+12)=0

We add all the numbers together, and all the variables

-(-2c(-2c+12)+12)+6=0


We calculate terms in parentheses: -(-2c(-2c+12)+12), so:

-2c(-2c+12)+12

We multiply parentheses

4c^2-24c+12

Back to the equation:

-(4c^2-24c+12)


We get rid of parentheses

-4c^2+24c-12+6=0

We add all the numbers together, and all the variables

-4c^2+24c-6=0

a = -4; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·(-4)·(-6)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{30}}{2*-4}=\frac{-24-4\sqrt{30}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{30}}{2*-4}=\frac{-24+4\sqrt{30}}{-8}
$