(3c+6)(3c+6)=64

Solution for (3c+6)(3c+6)=64 equation:

(3c+6)(3c+6)=64

We move all terms to the left:

(3c+6)(3c+6)-(64)=0

We multiply parentheses ..

(+9c^2+18c+18c+36)-64=0

We get rid of parentheses

9c^2+18c+18c+36-64=0

We add all the numbers together, and all the variables

9c^2+36c-28=0

a = 9; b = 36; c = -28;
Δ = b2-4ac
Δ = 362-4·9·(-28)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-48}{2*9}=\frac{-84}{18}
=-4+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+48}{2*9}=\frac{12}{18}
=2/3
$