3c(c+2)=0

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

3c(c+2)=0

We multiply parentheses

3c^2+6c=0

a = 3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·3·0
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{-(6)-6}{2*3}=\frac{-12}{6}
=-2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*3}=\frac{0}{6}
=0
$