2k^2-2k-3=0

Solution for 2k^2-2k-3=0 equation:

2k^2-2k-3=0

a = 2; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·2·(-3)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*2}=\frac{2-2\sqrt{7}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*2}=\frac{2+2\sqrt{7}}{4}
$