F(n)=2/3n+7

Solution for F(n)=2/3n+7 equation:

(F)=2/3F+7

We move all terms to the left:

(F)-(2/3F+7)=0


Domain of the equation: 3F+7)!=0

F∈R


We get rid of parentheses

F-2/3F-7=0

We multiply all the terms by the denominator


F*3F-7*3F-2=0

Wy multiply elements

3F^2-21F-2=0

a = 3; b = -21; c = -2;
Δ = b2-4ac
Δ = -212-4·3·(-2)
Δ = 465
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{-(-21)-\sqrt{465}}{2*3}=\frac{21-\sqrt{465}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{465}}{2*3}=\frac{21+\sqrt{465}}{6}
$