A=(4x-2)(3x+3)

Solution for A=(4x-2)(3x+3) equation:

=(4A-2)(3A+3)

We move all terms to the left:

-((4A-2)(3A+3))=0

We multiply parentheses ..

-((+12A^2+12A-6A-6))=0


We calculate terms in parentheses: -((+12A^2+12A-6A-6)), so:

(+12A^2+12A-6A-6)

We get rid of parentheses

12A^2+12A-6A-6

We add all the numbers together, and all the variables

12A^2+6A-6

Back to the equation:

-(12A^2+6A-6)


We get rid of parentheses

-12A^2-6A+6=0

a = -12; b = -6; c = +6;
Δ = b2-4ac
Δ = -62-4·(-12)·6
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-12}=\frac{-12}{-24}
=1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-12}=\frac{24}{-24}
=-1
$