A=(3x+5)(6x-4)

Solution for A=(3x+5)(6x-4) equation:

=(3A+5)(6A-4)

We move all terms to the left:

-((3A+5)(6A-4))=0

We multiply parentheses ..

-((+18A^2-12A+30A-20))=0


We calculate terms in parentheses: -((+18A^2-12A+30A-20)), so:

(+18A^2-12A+30A-20)

We get rid of parentheses

18A^2-12A+30A-20

We add all the numbers together, and all the variables

18A^2+18A-20

Back to the equation:

-(18A^2+18A-20)


We get rid of parentheses

-18A^2-18A+20=0

a = -18; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·(-18)·20
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-42}{2*-18}=\frac{-24}{-36}
=2/3
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+42}{2*-18}=\frac{60}{-36}
=-1+2/3
$