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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+2A^2+10A-9A-45))=0


We calculate terms in parentheses: -((+2A^2+10A-9A-45)), so:

(+2A^2+10A-9A-45)

We get rid of parentheses

2A^2+10A-9A-45

We add all the numbers together, and all the variables

2A^2+A-45

Back to the equation:

-(2A^2+A-45)


We get rid of parentheses

-2A^2-A+45=0

We add all the numbers together, and all the variables

-2A^2-1A+45=0

a = -2; b = -1; c = +45;
Δ = b2-4ac
Δ = -12-4·(-2)·45
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-19}{2*-2}=\frac{-18}{-4}
=4+1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+19}{2*-2}=\frac{20}{-4}
=-5
$