2x2-8x-8=0

Solution for 2x2-8x-8=0 equation:

2x^2-8x-8=0

a = 2; b = -8; c = -8;
Δ = b2-4ac
Δ = -82-4·2·(-8)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*2}=\frac{8-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*2}=\frac{8+8\sqrt{2}}{4}
$