A=(2x+9)(4x+1)

Solution for A=(2x+9)(4x+1) equation:

=(2A+9)(4A+1)

We move all terms to the left:

-((2A+9)(4A+1))=0

We multiply parentheses ..

-((+8A^2+2A+36A+9))=0


We calculate terms in parentheses: -((+8A^2+2A+36A+9)), so:

(+8A^2+2A+36A+9)

We get rid of parentheses

8A^2+2A+36A+9

We add all the numbers together, and all the variables

8A^2+38A+9

Back to the equation:

-(8A^2+38A+9)


We get rid of parentheses

-8A^2-38A-9=0

a = -8; b = -38; c = -9;
Δ = b2-4ac
Δ = -382-4·(-8)·(-9)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-34}{2*-8}=\frac{4}{-16}
=-1/4
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+34}{2*-8}=\frac{72}{-16}
=-4+1/2
$