2n(n-3)=0

Solution for 2n(n-3)=0 equation:

2n(n-3)=0

We multiply parentheses

2n^2-6n=0

a = 2; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·2·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*2}=\frac{0}{4}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*2}=\frac{12}{4}
=3
$