A=x^2+40x+x^2

Solution for A=x^2+40x+x^2 equation:

=A^2+40A+A^2

We move all terms to the left:

-(A^2+40A+A^2)=0

We get rid of parentheses

-A^2-A^2-40A=0

We add all the numbers together, and all the variables

-2A^2-40A=0

a = -2; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-2)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-2}=\frac{0}{-4}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-2}=\frac{80}{-4}
=-20
$