A=(5x+4)(2x-5)

Solution for A=(5x+4)(2x-5) equation:

=(5A+4)(2A-5)

We move all terms to the left:

-((5A+4)(2A-5))=0

We multiply parentheses ..

-((+10A^2-25A+8A-20))=0


We calculate terms in parentheses: -((+10A^2-25A+8A-20)), so:

(+10A^2-25A+8A-20)

We get rid of parentheses

10A^2-25A+8A-20

We add all the numbers together, and all the variables

10A^2-17A-20

Back to the equation:

-(10A^2-17A-20)


We get rid of parentheses

-10A^2+17A+20=0

a = -10; b = 17; c = +20;
Δ = b2-4ac
Δ = 172-4·(-10)·20
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-33}{2*-10}=\frac{-50}{-20}
=2+1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+33}{2*-10}=\frac{16}{-20}
=-4/5
$