S(n)=3/4n+1

Solution for S(n)=3/4n+1 equation:

(S)=3/4S+1

We move all terms to the left:

(S)-(3/4S+1)=0


Domain of the equation: 4S+1)!=0

S∈R


We get rid of parentheses

S-3/4S-1=0

We multiply all the terms by the denominator


S*4S-1*4S-3=0

Wy multiply elements

4S^2-4S-3=0

a = 4; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·4·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*4}=\frac{-4}{8}
=-1/2
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*4}=\frac{12}{8}
=1+1/2
$