F(n)=3+5/3n

Solution for F(n)=3+5/3n equation:

(F)=3+5/3F

We move all terms to the left:

(F)-(3+5/3F)=0


Domain of the equation: 3F)!=0

F!=0/1

F!=0

F∈R


We add all the numbers together, and all the variables

F-(5/3F+3)=0

We get rid of parentheses

F-5/3F-3=0

We multiply all the terms by the denominator


F*3F-3*3F-5=0

Wy multiply elements

3F^2-9F-5=0

a = 3; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·3·(-5)
Δ = 141
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{-(-9)-\sqrt{141}}{2*3}=\frac{9-\sqrt{141}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{141}}{2*3}=\frac{9+\sqrt{141}}{6}
$