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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(2c-6)3c-30=0

We multiply parentheses

6c^2-18c-30=0

a = 6; b = -18; c = -30;
Δ = b2-4ac
Δ = -182-4·6·(-30)
Δ = 1044
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:

$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{29}}{2*6}=\frac{18-6\sqrt{29}}{12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{29}}{2*6}=\frac{18+6\sqrt{29}}{12}
$