2(k^2)+2k-2=0

Solution for 2(k^2)+2k-2=0 equation:

2(k^2)+2k-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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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