(3t)/(t+2)=H(t)

Solution for (3t)/(t+2)=H(t) equation:

(3t)/(t+2)=(t)

We move all terms to the left:

(3t)/(t+2)-((t))=0


Domain of the equation: (t+2)!=0

We move all terms containing t to the left, all other terms to the right

t!=-2

t∈R


determiningTheFunctionDomain
3t/(t+2)-t=0

We add all the numbers together, and all the variables

-1t+3t/(t+2)=0

We multiply all the terms by the denominator


-1t*(t+2)+3t=0

We add all the numbers together, and all the variables

3t-1t*(t+2)=0

We multiply parentheses

-t^2+3t-2t=0

We add all the numbers together, and all the variables

-1t^2+t=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-1}=\frac{-2}{-2}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-1}=\frac{0}{-2}
=0
$