8u2+8u+1=0

Solution for 8u2+8u+1=0 equation:

8u^2+8u+1=0

a = 8; b = 8; c = +1;
Δ = b2-4ac
Δ = 82-4·8·1
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2}}{2*8}=\frac{-8-4\sqrt{2}}{16}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2}}{2*8}=\frac{-8+4\sqrt{2}}{16}
$