F(n)=160+6n-0.1n^2

Solution for F(n)=160+6n-0.1n^2 equation:

(F)=160+6F-0.1F^2

We move all terms to the left:

(F)-(160+6F-0.1F^2)=0

We get rid of parentheses

0.1F^2-6F+F-160=0

We add all the numbers together, and all the variables

0.1F^2-5F-160=0

a = 0.1; b = -5; c = -160;
Δ = b2-4ac
Δ = -52-4·0.1·(-160)
Δ = 89
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{-(-5)-\sqrt{89}}{2*0.1}=\frac{5-\sqrt{89}}{0.2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{89}}{2*0.1}=\frac{5+\sqrt{89}}{0.2}
$