F(n)=2/3n-5

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

(F)=2/3F-5

We move all terms to the left:

(F)-(2/3F-5)=0


Domain of the equation: 3F-5)!=0

F∈R


We get rid of parentheses

F-2/3F+5=0

We multiply all the terms by the denominator


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

Wy multiply elements

3F^2+15F-2=0

a = 3; b = 15; c = -2;
Δ = b2-4ac
Δ = 152-4·3·(-2)
Δ = 249
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{-(15)-\sqrt{249}}{2*3}=\frac{-15-\sqrt{249}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{249}}{2*3}=\frac{-15+\sqrt{249}}{6}
$