3c-c(6-2c)=27

Solution for 3c-c(6-2c)=27 equation:

3c-c(6-2c)=27

We move all terms to the left:

3c-c(6-2c)-(27)=0

We add all the numbers together, and all the variables

3c-c(-2c+6)-27=0

We multiply parentheses

2c^2+3c-6c-27=0

We add all the numbers together, and all the variables

2c^2-3c-27=0

a = 2; b = -3; c = -27;
Δ = b2-4ac
Δ = -32-4·2·(-27)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*2}=\frac{-12}{4}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*2}=\frac{18}{4}
=4+1/2
$