A(x)=(18+2x)(12+2x)

Solution for A(x)=(18+2x)(12+2x) equation:

(A)=(18+2A)(12+2A)

We move all terms to the left:

(A)-((18+2A)(12+2A))=0

We add all the numbers together, and all the variables

A-((2A+18)(2A+12))=0

We multiply parentheses ..

-((+4A^2+24A+36A+216))+A=0


We calculate terms in parentheses: -((+4A^2+24A+36A+216)), so:

(+4A^2+24A+36A+216)

We get rid of parentheses

4A^2+24A+36A+216

We add all the numbers together, and all the variables

4A^2+60A+216

Back to the equation:

-(4A^2+60A+216)


We add all the numbers together, and all the variables

A-(4A^2+60A+216)=0

We get rid of parentheses

-4A^2+A-60A-216=0

We add all the numbers together, and all the variables

-4A^2-59A-216=0

a = -4; b = -59; c = -216;
Δ = b2-4ac
Δ = -592-4·(-4)·(-216)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-5}{2*-4}=\frac{54}{-8}
=-6+3/4
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+5}{2*-4}=\frac{64}{-8}
=-8
$