(2n-5)(n-4)=0

Solution for (2n-5)(n-4)=0 equation:

(2n-5)(n-4)=0

We multiply parentheses ..

(+2n^2-8n-5n+20)=0

We get rid of parentheses

2n^2-8n-5n+20=0

We add all the numbers together, and all the variables

2n^2-13n+20=0

a = 2; b = -13; c = +20;
Δ = b2-4ac
Δ = -132-4·2·20
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*2}=\frac{10}{4}
=2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*2}=\frac{16}{4}
=4
$