(3b+9)(b-2)=0

Solution for (3b+9)(b-2)=0 equation:

(3b+9)(b-2)=0

We multiply parentheses ..

(+3b^2-6b+9b-18)=0

We get rid of parentheses

3b^2-6b+9b-18=0

We add all the numbers together, and all the variables

3b^2+3b-18=0

a = 3; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·3·(-18)
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{-(3)-15}{2*3}=\frac{-18}{6}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*3}=\frac{12}{6}
=2
$