Y=40x+-2x^2

Solution for Y=40x+-2x^2 equation:

=40Y+-2Y^2

We move all terms to the left:

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

We use the square of the difference formula

-(40Y-2Y^2)=0

We get rid of parentheses

2Y^2-40Y=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:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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