8k^2+8k-4=0

Solution for 8k^2+8k-4=0 equation:

8k^2+8k-4=0

a = 8; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·8·(-4)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{3}}{2*8}=\frac{-8-8\sqrt{3}}{16}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{3}}{2*8}=\frac{-8+8\sqrt{3}}{16}
$