3b^2+8b-108=0

Solution for 3b^2+8b-108=0 equation:

3b^2+8b-108=0

a = 3; b = 8; c = -108;
Δ = b2-4ac
Δ = 82-4·3·(-108)
Δ = 1360
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:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{85}}{2*3}=\frac{-8-4\sqrt{85}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{85}}{2*3}=\frac{-8+4\sqrt{85}}{6}
$