3b^2+32b-128=0

Solution for 3b^2+32b-128=0 equation:

3b^2+32b-128=0

a = 3; b = 32; c = -128;
Δ = b2-4ac
Δ = 322-4·3·(-128)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{10}}{2*3}=\frac{-32-16\sqrt{10}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{10}}{2*3}=\frac{-32+16\sqrt{10}}{6}
$