(3c+3)(4c+2)=0

Solution for (3c+3)(4c+2)=0 equation:

(3c+3)(4c+2)=0

We multiply parentheses ..

(+12c^2+6c+12c+6)=0

We get rid of parentheses

12c^2+6c+12c+6=0

We add all the numbers together, and all the variables

12c^2+18c+6=0

a = 12; b = 18; c = +6;
Δ = b2-4ac
Δ = 182-4·12·6
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*12}=\frac{-24}{24}
=-1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*12}=\frac{-12}{24}
=-1/2
$