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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4b^2-6b+6b+9)=0

We get rid of parentheses

-4b^2-6b+6b+9=0

We add all the numbers together, and all the variables

-4b^2+9=0

a = -4; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-4)·9
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-4}=\frac{12}{-8}
=-1+1/2
$