F(x)=-2x^2+40x

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

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

We move all terms to the left:

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

We get rid of parentheses

2F^2-40F+F=0

We add all the numbers together, and all the variables

2F^2-39F=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:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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