2n^2+18n-20=0

Solution for 2n^2+18n-20=0 equation:

2n^2+18n-20=0

a = 2; b = 18; c = -20;
Δ = b2-4ac
Δ = 182-4·2·(-20)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*2}=\frac{-40}{4}
=-10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*2}=\frac{4}{4}
=1
$