(2b+3)(3b-)=0

Solution for (2b+3)(3b-)=0 equation:

(2b+3)(3b-)=0

We add all the numbers together, and all the variables

(2b+3)(+3b)=0

We multiply parentheses ..

(+6b^2+9b)=0

We get rid of parentheses

6b^2+9b=0

a = 6; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·6·0
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*6}=\frac{-18}{12}
=-1+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*6}=\frac{0}{12}
=0
$