z(3z+5)=0

Solution for z(3z+5)=0 equation:

z(3z+5)=0

We multiply parentheses

3z^2+5z=0

a = 3; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·3·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*3}=\frac{-10}{6}
=-1+2/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*3}=\frac{0}{6}
=0
$