c(3c+8)=0

Solution for c(3c+8)=0 equation:

c(3c+8)=0

We multiply parentheses

3c^2+8c=0

a = 3; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·3·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*3}=\frac{-16}{6}
=-2+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*3}=\frac{0}{6}
=0
$