T(n)=7T(n/2)+18n2

Solution for T(n)=7T(n/2)+18n2 equation:

(T)=7(T/2)+18T^2

We move all terms to the left:

(T)-(7(T/2)+18T^2)=0


Domain of the equation: 2)+18T^2)!=0

T!=0/1

T!=0

T∈R


We add all the numbers together, and all the variables

-(7(+T/2)+18T^2)+T=0

We multiply all the terms by the denominator


-(7(+T+T*2)+18T^2)=0


We calculate terms in parentheses: -(7(+T+T*2)+18T^2), so:

7(+T+T*2)+18T^2

determiningTheFunctionDomain
18T^2+7(+T+T*2)

We multiply parentheses

18T^2+7T+14T

We add all the numbers together, and all the variables

18T^2+21T

Back to the equation:

-(18T^2+21T)


We get rid of parentheses

-18T^2-21T=0

a = -18; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·(-18)·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*-18}=\frac{0}{-36}
=0
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*-18}=\frac{42}{-36}
=-1+1/6
$