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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(3c-c(-2c+6)), so:

3c-c(-2c+6)

We multiply parentheses

2c^2+3c-6c

We add all the numbers together, and all the variables

2c^2-3c

Back to the equation:

-(2c^2-3c)


We get rid of parentheses

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