2y^2-4y-4=0

Solution for 2y^2-4y-4=0 equation:

2y^2-4y-4=0

a = 2; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·2·(-4)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*2}=\frac{4-4\sqrt{3}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*2}=\frac{4+4\sqrt{3}}{4}
$