B3-8b2+16b=0

Solution for B3-8b2+16b=0 equation:

3-8B^2+16B=0

a = -8; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·(-8)·3
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{22}}{2*-8}=\frac{-16-4\sqrt{22}}{-16}
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{22}}{2*-8}=\frac{-16+4\sqrt{22}}{-16}
$