A=(150+3x)(300+3x)

Solution for A=(150+3x)(300+3x) equation:

=(150+3A)(300+3A)

We move all terms to the left:

-((150+3A)(300+3A))=0

We add all the numbers together, and all the variables

-((3A+150)(3A+300))=0

We multiply parentheses ..

-((+9A^2+900A+450A+45000))=0


We calculate terms in parentheses: -((+9A^2+900A+450A+45000)), so:

(+9A^2+900A+450A+45000)

We get rid of parentheses

9A^2+900A+450A+45000

We add all the numbers together, and all the variables

9A^2+1350A+45000

Back to the equation:

-(9A^2+1350A+45000)


We get rid of parentheses

-9A^2-1350A-45000=0

a = -9; b = -1350; c = -45000;
Δ = b2-4ac
Δ = -13502-4·(-9)·(-45000)
Δ = 202500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{202500}=450$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1350)-450}{2*-9}=\frac{900}{-18}
=-50
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1350)+450}{2*-9}=\frac{1800}{-18}
=-100
$