t2/3;t=2/4

Solution for t2/3;t=2/4 equation:

t2/3t=2/4

We move all terms to the left:

t2/3t-(2/4)=0


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We add all the numbers together, and all the variables

t2/3t-(+2/4)=0

We get rid of parentheses

t2/3t-2/4=0

We calculate fractions


4t^2/12t+(-6t)/12t=0

We multiply all the terms by the denominator


4t^2+(-6t)=0

We get rid of parentheses

4t^2-6t=0

a = 4; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·4·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*4}=\frac{0}{8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*4}=\frac{12}{8}
=1+1/2
$