(n+2)(n-1)=255

Solution for (n+2)(n-1)=255 equation:

(n+2)(n-1)=255

We move all terms to the left:

(n+2)(n-1)-(255)=0

We multiply parentheses ..

(+n^2-1n+2n-2)-255=0

We get rid of parentheses

n^2-1n+2n-2-255=0

We add all the numbers together, and all the variables

n^2+n-257=0

a = 1; b = 1; c = -257;
Δ = b2-4ac
Δ = 12-4·1·(-257)
Δ = 1029
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:

$\sqrt{\Delta}=\sqrt{1029}=\sqrt{49*21}=\sqrt{49}*\sqrt{21}=7\sqrt{21}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7\sqrt{21}}{2*1}=\frac{-1-7\sqrt{21}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7\sqrt{21}}{2*1}=\frac{-1+7\sqrt{21}}{2}
$