A(x)=8x^2+20x

Solution for A(x)=8x^2+20x equation:

(A)=8A^2+20A

We move all terms to the left:

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

We get rid of parentheses

-8A^2+A-20A=0

We add all the numbers together, and all the variables

-8A^2-19A=0

a = -8; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·(-8)·0
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{-(-19)-19}{2*-8}=\frac{0}{-16}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*-8}=\frac{38}{-16}
=-2+3/8
$