2n(n-3)+1=2n-5

Solution for 2n(n-3)+1=2n-5 equation:

2n(n-3)+1=2n-5

We move all terms to the left:

2n(n-3)+1-(2n-5)=0

We multiply parentheses

2n^2-6n-(2n-5)+1=0

We get rid of parentheses

2n^2-6n-2n+5+1=0

We add all the numbers together, and all the variables

2n^2-8n+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$