F(n)=2+5/3n

Solution for F(n)=2+5/3n equation:

(F)=2+5/3F

We move all terms to the left:

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

We get rid of parentheses

F-5/3F-2=0

We multiply all the terms by the denominator


F*3F-2*3F-5=0

Wy multiply elements

3F^2-6F-5=0

a = 3; b = -6; c = -5;
Δ = b2-4ac
Δ = -62-4·3·(-5)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{6}}{2*3}=\frac{6-4\sqrt{6}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{6}}{2*3}=\frac{6+4\sqrt{6}}{6}
$