2n^2-20=18n

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

2n^2-20=18n

We move all terms to the left:

2n^2-20-(18n)=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{-4}{4}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*2}=\frac{40}{4}
=10
$