(6b+3)(2-b)=0

Solution for (6b+3)(2-b)=0 equation:

(6b+3)(2-b)=0

We add all the numbers together, and all the variables

(6b+3)(-1b+2)=0

We multiply parentheses ..

(-6b^2+12b-3b+6)=0

We get rid of parentheses

-6b^2+12b-3b+6=0

We add all the numbers together, and all the variables

-6b^2+9b+6=0

a = -6; b = 9; c = +6;
Δ = b2-4ac
Δ = 92-4·(-6)·6
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*-6}=\frac{-24}{-12}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*-6}=\frac{6}{-12}
=-1/2
$