1/3t-3=-2t+4

Solution for 1/3t-3=-2t+4 equation:

1/3t-3=-2t+4

We move all terms to the left:

1/3t-3-(-2t+4)=0


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We get rid of parentheses

1/3t+2t-4-3=0

We multiply all the terms by the denominator


2t*3t-4*3t-3*3t+1=0

Wy multiply elements

6t^2-12t-9t+1=0

We add all the numbers together, and all the variables

6t^2-21t+1=0

a = 6; b = -21; c = +1;
Δ = b2-4ac
Δ = -212-4·6·1
Δ = 417
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{417}}{2*6}=\frac{21-\sqrt{417}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{417}}{2*6}=\frac{21+\sqrt{417}}{12}
$