3b(4b-3)=0

Solution for 3b(4b-3)=0 equation:

3b(4b-3)=0

We multiply parentheses

12b^2-9b=0

a = 12; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·12·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*12}=\frac{0}{24}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*12}=\frac{18}{24}
=3/4
$