n^2+19n-20=0

Solution for n^2+19n-20=0 equation:

n^2+19n-20=0

a = 1; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·1·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*1}=\frac{-40}{2}
=-20
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*1}=\frac{2}{2}
=1
$