A(x)=-2x^2+40X

Solution for A(x)=-2x^2+40X equation:

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

We move all terms to the left:

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

We get rid of parentheses

2A^2-40A+A=0

We add all the numbers together, and all the variables

2A^2-39A=0

a = 2; b = -39; c = 0;
Δ = b2-4ac
Δ = -392-4·2·0
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-39}{2*2}=\frac{0}{4}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+39}{2*2}=\frac{78}{4}
=19+1/2
$