3c^2+6=27-3c

Solution for 3c^2+6=27-3c equation:

3c^2+6=27-3c

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3c^2+3c-21=0

a = 3; b = 3; c = -21;
Δ = b2-4ac
Δ = 32-4·3·(-21)
Δ = 261
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:

$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{29}}{2*3}=\frac{-3-3\sqrt{29}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{29}}{2*3}=\frac{-3+3\sqrt{29}}{6}
$