(n+2)(n+1)=240

Solution for (n+2)(n+1)=240 equation:

(n+2)(n+1)=240

We move all terms to the left:

(n+2)(n+1)-(240)=0

We multiply parentheses ..

(+n^2+n+2n+2)-240=0

We get rid of parentheses

n^2+n+2n+2-240=0

We add all the numbers together, and all the variables

n^2+3n-238=0

a = 1; b = 3; c = -238;
Δ = b2-4ac
Δ = 32-4·1·(-238)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-31}{2*1}=\frac{-34}{2}
=-17
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+31}{2*1}=\frac{28}{2}
=14
$