(2n-3)(n-1)=0

Solution for (2n-3)(n-1)=0 equation:

(2n-3)(n-1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-5n+3=0

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

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*2}=\frac{4}{4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*2}=\frac{6}{4}
=1+1/2
$