(3n-2)(n-3)=0

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

(3n-2)(n-3)=0

We multiply parentheses ..

(+3n^2-9n-2n+6)=0

We get rid of parentheses

3n^2-9n-2n+6=0

We add all the numbers together, and all the variables

3n^2-11n+6=0

a = 3; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·3·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*3}=\frac{4}{6}
=2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*3}=\frac{18}{6}
=3
$