F(n)=1/3n+16

Solution for F(n)=1/3n+16 equation:

(F)=1/3F+16

We move all terms to the left:

(F)-(1/3F+16)=0


Domain of the equation: 3F+16)!=0

F∈R


We get rid of parentheses

F-1/3F-16=0

We multiply all the terms by the denominator


F*3F-16*3F-1=0

Wy multiply elements

3F^2-48F-1=0

a = 3; b = -48; c = -1;
Δ = b2-4ac
Δ = -482-4·3·(-1)
Δ = 2316
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:

$\sqrt{\Delta}=\sqrt{2316}=\sqrt{4*579}=\sqrt{4}*\sqrt{579}=2\sqrt{579}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-2\sqrt{579}}{2*3}=\frac{48-2\sqrt{579}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+2\sqrt{579}}{2*3}=\frac{48+2\sqrt{579}}{6}
$