n^2+3n+2=1440

Solution for n^2+3n+2=1440 equation:

n^2+3n+2=1440

We move all terms to the left:

n^2+3n+2-(1440)=0

We add all the numbers together, and all the variables

n^2+3n-1438=0

a = 1; b = 3; c = -1438;
Δ = b2-4ac
Δ = 32-4·1·(-1438)
Δ = 5761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{5761}}{2*1}=\frac{-3-\sqrt{5761}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5761}}{2*1}=\frac{-3+\sqrt{5761}}{2}
$