2n(n+2)=4n+1-2n

Solution for 2n(n+2)=4n+1-2n equation:

2n(n+2)=4n+1-2n

We move all terms to the left:

2n(n+2)-(4n+1-2n)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

2n^2+4n-(2n+1)=0

We get rid of parentheses

2n^2+4n-2n-1=0

We add all the numbers together, and all the variables

2n^2+2n-1=0

a = 2; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·2·(-1)
Δ = 12
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:

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