c(3c+1)=0

Solution for c(3c+1)=0 equation:

c(3c+1)=0

We multiply parentheses

3c^2+c=0

a = 3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·3·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*3}=\frac{-2}{6}
=-1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*3}=\frac{0}{6}
=0
$