8u^2+8u-2=0

Solution for 8u^2+8u-2=0 equation:

8u^2+8u-2=0

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

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*8}=\frac{-8-8\sqrt{2}}{16}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*8}=\frac{-8+8\sqrt{2}}{16}
$