2y^2-16y-8=0

Solution for 2y^2-16y-8=0 equation:

2y^2-16y-8=0

a = 2; b = -16; c = -8;
Δ = b2-4ac
Δ = -162-4·2·(-8)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{5}}{2*2}=\frac{16-8\sqrt{5}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{5}}{2*2}=\frac{16+8\sqrt{5}}{4}
$