F(n)=-5n^2+400n-6000

Solution for F(n)=-5n^2+400n-6000 equation:

(F)=-5F^2+400F-6000

We move all terms to the left:

(F)-(-5F^2+400F-6000)=0

We get rid of parentheses

5F^2-400F+F+6000=0

We add all the numbers together, and all the variables

5F^2-399F+6000=0

a = 5; b = -399; c = +6000;
Δ = b2-4ac
Δ = -3992-4·5·6000
Δ = 39201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-399)-\sqrt{39201}}{2*5}=\frac{399-\sqrt{39201}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-399)+\sqrt{39201}}{2*5}=\frac{399+\sqrt{39201}}{10}
$