A=(3x+5)(2x-4)

Solution for A=(3x+5)(2x-4) equation:

=(3A+5)(2A-4)

We move all terms to the left:

-((3A+5)(2A-4))=0

We multiply parentheses ..

-((+6A^2-12A+10A-20))=0


We calculate terms in parentheses: -((+6A^2-12A+10A-20)), so:

(+6A^2-12A+10A-20)

We get rid of parentheses

6A^2-12A+10A-20

We add all the numbers together, and all the variables

6A^2-2A-20

Back to the equation:

-(6A^2-2A-20)


We get rid of parentheses

-6A^2+2A+20=0

a = -6; b = 2; c = +20;
Δ = b2-4ac
Δ = 22-4·(-6)·20
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*-6}=\frac{-24}{-12}
=+2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*-6}=\frac{20}{-12}
=-1+2/3
$