A=(30+2x)(40+2x)

Solution for A=(30+2x)(40+2x) equation:

=(30+2A)(40+2A)

We move all terms to the left:

-((30+2A)(40+2A))=0

We add all the numbers together, and all the variables

-((2A+30)(2A+40))=0

We multiply parentheses ..

-((+4A^2+80A+60A+1200))=0


We calculate terms in parentheses: -((+4A^2+80A+60A+1200)), so:

(+4A^2+80A+60A+1200)

We get rid of parentheses

4A^2+80A+60A+1200

We add all the numbers together, and all the variables

4A^2+140A+1200

Back to the equation:

-(4A^2+140A+1200)


We get rid of parentheses

-4A^2-140A-1200=0

a = -4; b = -140; c = -1200;
Δ = b2-4ac
Δ = -1402-4·(-4)·(-1200)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-20}{2*-4}=\frac{120}{-8}
=-15
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+20}{2*-4}=\frac{160}{-8}
=-20
$