T(n)=T(5n/3)+C1

Solution for T(n)=T(5n/3)+C1 equation:

(T)=(5T/3)+1

We move all terms to the left:

(T)-((5T/3)+1)=0

We add all the numbers together, and all the variables

T-((+5T/3)+1)=0

We multiply all the terms by the denominator


T*3)+1)-((+5T=0

We add all the numbers together, and all the variables

5T+T*3)+1)-((=0

Wy multiply elements

3T^2+5T=0

a = 3; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·3·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*3}=\frac{-10}{6}
=-1+2/3
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*3}=\frac{0}{6}
=0
$