2n^2-4n-440=0

Solution for 2n^2-4n-440=0 equation:

2n^2-4n-440=0

a = 2; b = -4; c = -440;
Δ = b2-4ac
Δ = -42-4·2·(-440)
Δ = 3536
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:

$\sqrt{\Delta}=\sqrt{3536}=\sqrt{16*221}=\sqrt{16}*\sqrt{221}=4\sqrt{221}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{221}}{2*2}=\frac{4-4\sqrt{221}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{221}}{2*2}=\frac{4+4\sqrt{221}}{4}
$