A=(2x+5)(x-9)

Solution for A=(2x+5)(x-9) equation:

=(2A+5)(A-9)

We move all terms to the left:

-((2A+5)(A-9))=0

We multiply parentheses ..

-((+2A^2-18A+5A-45))=0


We calculate terms in parentheses: -((+2A^2-18A+5A-45)), so:

(+2A^2-18A+5A-45)

We get rid of parentheses

2A^2-18A+5A-45

We add all the numbers together, and all the variables

2A^2-13A-45

Back to the equation:

-(2A^2-13A-45)


We get rid of parentheses

-2A^2+13A+45=0

a = -2; b = 13; c = +45;
Δ = b2-4ac
Δ = 132-4·(-2)·45
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*-2}=\frac{-36}{-4}
=+9
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*-2}=\frac{10}{-4}
=-2+1/2
$