n2+4n-480=0

Solution for n2+4n-480=0 equation:

n2+4n-480=0

We add all the numbers together, and all the variables

n^2+4n-480=0

a = 1; b = 4; c = -480;
Δ = b2-4ac
Δ = 42-4·1·(-480)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*1}=\frac{-48}{2}
=-24
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*1}=\frac{40}{2}
=20
$