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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((5A+4)+(+10A^2-35A+8A-28))=0


We calculate terms in parentheses: -((5A+4)+(+10A^2-35A+8A-28)), so:

(5A+4)+(+10A^2-35A+8A-28)

determiningTheFunctionDomain
(+10A^2-35A+8A-28)+(5A+4)

We get rid of parentheses

10A^2-35A+8A+5A-28+4

We add all the numbers together, and all the variables

10A^2-22A-24

Back to the equation:

-(10A^2-22A-24)


We get rid of parentheses

-10A^2+22A+24=0

a = -10; b = 22; c = +24;
Δ = b2-4ac
Δ = 222-4·(-10)·24
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-38}{2*-10}=\frac{-60}{-20}
=+3
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*-10}=\frac{16}{-20}
=-4/5
$