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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+8A^2+18A+4A+9))=0


We calculate terms in parentheses: -((+8A^2+18A+4A+9)), so:

(+8A^2+18A+4A+9)

We get rid of parentheses

8A^2+18A+4A+9

We add all the numbers together, and all the variables

8A^2+22A+9

Back to the equation:

-(8A^2+22A+9)


We get rid of parentheses

-8A^2-22A-9=0

a = -8; b = -22; c = -9;
Δ = b2-4ac
Δ = -222-4·(-8)·(-9)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*-8}=\frac{8}{-16}
=-1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*-8}=\frac{36}{-16}
=-2+1/4
$