n(2)-3-n+4=n+2

Solution for n(2)-3-n+4=n+2 equation:

n(2)-3-n+4=n+2

We move all terms to the left:

n(2)-3-n+4-(n+2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^2-2n-1=0

a = 1; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·1·(-1)
Δ = 8
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:

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