F(n)=(25n-n^2)/3

Solution for F(n)=(25n-n^2)/3 equation:

(F)=(25F-F^2)/3

We move all terms to the left:

(F)-((25F-F^2)/3)=0

We multiply all the terms by the denominator


-((25F-F^2)+F*3)=0


We calculate terms in parentheses: -((25F-F^2)+F*3), so:

(25F-F^2)+F*3

Wy multiply elements

(25F-F^2)+3F

We get rid of parentheses

-F^2+25F+3F

We add all the numbers together, and all the variables

-1F^2+28F

Back to the equation:

-(-1F^2+28F)


We get rid of parentheses

1F^2-28F=0

We add all the numbers together, and all the variables

F^2-28F=0

a = 1; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·1·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*1}=\frac{0}{2}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*1}=\frac{56}{2}
=28
$