2n^2-2n-2=0

Solution for 2n^2-2n-2=0 equation:

2n^2-2n-2=0

a = 2; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·2·(-2)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*2}=\frac{2-2\sqrt{5}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*2}=\frac{2+2\sqrt{5}}{4}
$