1/(6t-4)=3t-2

Solution for 1/(6t-4)=3t-2 equation:

1/(6t-4)=3t-2

We move all terms to the left:

1/(6t-4)-(3t-2)=0


Domain of the equation: (6t-4)!=0

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

6t!=4

t!=4/6

t!=2/3

t∈R


We get rid of parentheses

1/(6t-4)-3t+2=0

We multiply all the terms by the denominator


-3t*(6t-4)+2*(6t-4)+1=0

We multiply parentheses

-18t^2+12t+12t-8+1=0

We add all the numbers together, and all the variables

-18t^2+24t-7=0

a = -18; b = 24; c = -7;
Δ = b2-4ac
Δ = 242-4·(-18)·(-7)
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{2}}{2*-18}=\frac{-24-6\sqrt{2}}{-36}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{2}}{2*-18}=\frac{-24+6\sqrt{2}}{-36}
$