A=(12x)(4x+9)

Solution for A=(12x)(4x+9) equation:

=(12A)(4A+9)

We move all terms to the left:

-((12A)(4A+9))=0


We calculate terms in parentheses: -(12A(4A+9)), so:

12A(4A+9)

We multiply parentheses

48A^2+108A

Back to the equation:

-(48A^2+108A)


We get rid of parentheses

-48A^2-108A=0

a = -48; b = -108; c = 0;
Δ = b2-4ac
Δ = -1082-4·(-48)·0
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-108}{2*-48}=\frac{0}{-96}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+108}{2*-48}=\frac{216}{-96}
=-2+1/4
$