A=(3+4b)(3-8b)

Solution for A=(3+4b)(3-8b) equation:

=(3+4A)(3-8A)

We move all terms to the left:

-((3+4A)(3-8A))=0

We add all the numbers together, and all the variables

-((4A+3)(-8A+3))=0

We multiply parentheses ..

-((-32A^2+12A-24A+9))=0


We calculate terms in parentheses: -((-32A^2+12A-24A+9)), so:

(-32A^2+12A-24A+9)

We get rid of parentheses

-32A^2+12A-24A+9

We add all the numbers together, and all the variables

-32A^2-12A+9

Back to the equation:

-(-32A^2-12A+9)


We get rid of parentheses

32A^2+12A-9=0

a = 32; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·32·(-9)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*32}=\frac{-48}{64}
=-3/4
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*32}=\frac{24}{64}
=3/8
$