F(n)=3/2n-4

Solution for F(n)=3/2n-4 equation:

(F)=3/2F-4

We move all terms to the left:

(F)-(3/2F-4)=0


Domain of the equation: 2F-4)!=0

F∈R


We get rid of parentheses

F-3/2F+4=0

We multiply all the terms by the denominator


F*2F+4*2F-3=0

Wy multiply elements

2F^2+8F-3=0

a = 2; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·2·(-3)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{22}}{2*2}=\frac{-8-2\sqrt{22}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{22}}{2*2}=\frac{-8+2\sqrt{22}}{4}
$