2b^2+8b-2=0

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

2b^2+8b-2=0

a = 2; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·2·(-2)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{5}}{2*2}=\frac{-8-4\sqrt{5}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{5}}{2*2}=\frac{-8+4\sqrt{5}}{4}
$