1/(3x)-5/(6x)=17

Solution for 1/(3x)-5/(6x)=17 equation:

1/(3x)-5/(6x)=17

We move all terms to the left:

1/(3x)-5/(6x)-(17)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 6x!=0

x!=0/6

x!=0

x∈R


We calculate fractions


6x/18x^2+(-15x)/18x^2-17=0

We multiply all the terms by the denominator


6x+(-15x)-17*18x^2=0

Wy multiply elements

-306x^2+6x+(-15x)=0

We get rid of parentheses

-306x^2+6x-15x=0

We add all the numbers together, and all the variables

-306x^2-9x=0

a = -306; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-306)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-306}=\frac{0}{-612}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-306}=\frac{18}{-612}
=-1/34
$