2n+20=22n^2

Solution for 2n+20=22n^2 equation:

2n+20=22n^2

We move all terms to the left:

2n+20-(22n^2)=0

determiningTheFunctionDomain
-22n^2+2n+20=0

a = -22; b = 2; c = +20;
Δ = b2-4ac
Δ = 22-4·(-22)·20
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-42}{2*-22}=\frac{-44}{-44}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+42}{2*-22}=\frac{40}{-44}
=-10/11
$