(n+1)(n+1)=1440

Solution for (n+1)(n+1)=1440 equation:

(n+1)(n+1)=1440

We move all terms to the left:

(n+1)(n+1)-(1440)=0

We multiply parentheses ..

(+n^2+n+n+1)-1440=0

We get rid of parentheses

n^2+n+n+1-1440=0

We add all the numbers together, and all the variables

n^2+2n-1439=0

a = 1; b = 2; c = -1439;
Δ = b2-4ac
Δ = 22-4·1·(-1439)
Δ = 5760
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:

$\sqrt{\Delta}=\sqrt{5760}=\sqrt{576*10}=\sqrt{576}*\sqrt{10}=24\sqrt{10}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-24\sqrt{10}}{2*1}=\frac{-2-24\sqrt{10}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+24\sqrt{10}}{2*1}=\frac{-2+24\sqrt{10}}{2}
$