2n^2-2n=22

Solution for 2n^2-2n=22 equation:

2n^2-2n=22

We move all terms to the left:

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

a = 2; b = -2; c = -22;
Δ = b2-4ac
Δ = -22-4·2·(-22)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{5}}{2*2}=\frac{2-6\sqrt{5}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{5}}{2*2}=\frac{2+6\sqrt{5}}{4}
$