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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2n^2-6n-(2n-2)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-8n+2=0

a = 2; b = -8; c = +2;
Δ = b2-4ac
Δ = -82-4·2·2
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*2}=\frac{8-4\sqrt{3}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*2}=\frac{8+4\sqrt{3}}{4}
$