A=(3x-7)(4x+5)+(3x-7)2

Solution for A=(3x-7)(4x+5)+(3x-7)2 equation:

=(3A-7)(4A+5)+(3A-7)2

We move all terms to the left:

-((3A-7)(4A+5)+(3A-7)2)=0

We multiply parentheses ..

-((+12A^2+15A-28A-35)+(3A-7)2)=0


We calculate terms in parentheses: -((+12A^2+15A-28A-35)+(3A-7)2), so:

(+12A^2+15A-28A-35)+(3A-7)2

We multiply parentheses

(+12A^2+15A-28A-35)+6A-14

We get rid of parentheses

12A^2+15A-28A+6A-35-14

We add all the numbers together, and all the variables

12A^2-7A-49

Back to the equation:

-(12A^2-7A-49)


We get rid of parentheses

-12A^2+7A+49=0

a = -12; b = 7; c = +49;
Δ = b2-4ac
Δ = 72-4·(-12)·49
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-49}{2*-12}=\frac{-56}{-24}
=2+1/3
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+49}{2*-12}=\frac{42}{-24}
=-1+3/4
$