A=(7x+9)(4+x)

Solution for A=(7x+9)(4+x) equation:

=(7A+9)(4+A)

We move all terms to the left:

-((7A+9)(4+A))=0

We add all the numbers together, and all the variables

-((7A+9)(A+4))=0

We multiply parentheses ..

-((+7A^2+28A+9A+36))=0


We calculate terms in parentheses: -((+7A^2+28A+9A+36)), so:

(+7A^2+28A+9A+36)

We get rid of parentheses

7A^2+28A+9A+36

We add all the numbers together, and all the variables

7A^2+37A+36

Back to the equation:

-(7A^2+37A+36)


We get rid of parentheses

-7A^2-37A-36=0

a = -7; b = -37; c = -36;
Δ = b2-4ac
Δ = -372-4·(-7)·(-36)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-37)-19}{2*-7}=\frac{18}{-14}
=-1+2/7
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-37)+19}{2*-7}=\frac{56}{-14}
=-4
$