b^2-8b+2=0

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

b^2-8b+2=0

a = 1; b = -8; c = +2;
Δ = b2-4ac
Δ = -82-4·1·2
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{14}}{2*1}=\frac{8-2\sqrt{14}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{14}}{2*1}=\frac{8+2\sqrt{14}}{2}
$