(c2-6)3-3c=27

Solution for (c2-6)3-3c=27 equation:

(c2-6)3-3c=27

We move all terms to the left:

(c2-6)3-3c-(27)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

3c^2-3c-18-27=0

We add all the numbers together, and all the variables

3c^2-3c-45=0

a = 3; b = -3; c = -45;
Δ = b2-4ac
Δ = -32-4·3·(-45)
Δ = 549
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:

$\sqrt{\Delta}=\sqrt{549}=\sqrt{9*61}=\sqrt{9}*\sqrt{61}=3\sqrt{61}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{61}}{2*3}=\frac{3-3\sqrt{61}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{61}}{2*3}=\frac{3+3\sqrt{61}}{6}
$