A=(3x+1)(4x-1)

Solution for A=(3x+1)(4x-1) equation:

=(3A+1)(4A-1)

We move all terms to the left:

-((3A+1)(4A-1))=0

We multiply parentheses ..

-((+12A^2-3A+4A-1))=0


We calculate terms in parentheses: -((+12A^2-3A+4A-1)), so:

(+12A^2-3A+4A-1)

We get rid of parentheses

12A^2-3A+4A-1

We add all the numbers together, and all the variables

12A^2+A-1

Back to the equation:

-(12A^2+A-1)


We get rid of parentheses

-12A^2-A+1=0

We add all the numbers together, and all the variables

-12A^2-1A+1=0

a = -12; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·(-12)·1
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-12}=\frac{-6}{-24}
=1/4
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-12}=\frac{8}{-24}
=-1/3
$