2n^2+8n-2040=0

Solution for 2n^2+8n-2040=0 equation:

2n^2+8n-2040=0

a = 2; b = 8; c = -2040;
Δ = b2-4ac
Δ = 82-4·2·(-2040)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-128}{2*2}=\frac{-136}{4}
=-34
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+128}{2*2}=\frac{120}{4}
=30
$