A=(x+1)(x-2)(x+3)(x-3)

Solution for A=(x+1)(x-2)(x+3)(x-3) equation:

=(A+1)(A-2)(A+3)(A-3)

We move all terms to the left:

-((A+1)(A-2)(A+3)(A-3))=0

We multiply parentheses ..

-((+A^2-2A+A-2)(A+3)(A-3))=0


We calculate terms in parentheses: -((+A^2-2A+A-2)(A+3)(A-3)), so:

(+A^2-2A+A-2)(A+3)(A-3)

We use the square of the difference formula

A^2-9

Back to the equation:

-(A^2-9)


We get rid of parentheses

-A^2+9=0

We add all the numbers together, and all the variables

-1A^2+9=0

a = -1; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-1)·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-1}=\frac{-6}{-2}
=+3
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-1}=\frac{6}{-2}
=-3
$