5/(3x+2)+(3/x-4)=0

Solution for 5/(3x+2)+(3/x-4)=0 equation:

5/(3x+2)+(3/x-4)=0


Domain of the equation: (3x+2)!=0

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

3x!=-2

x!=-2/3

x!=-2/3

x∈R



Domain of the equation: x-4)!=0

x∈R


We get rid of parentheses

5/(3x+2)+3/x-4=0

We calculate fractions


5x/(3x^2+2x)+(9x+6)/(3x^2+2x)-4=0

We multiply all the terms by the denominator


5x+(9x+6)-4*(3x^2+2x)=0

We multiply parentheses

-12x^2+5x+(9x+6)-8x=0

We get rid of parentheses

-12x^2+5x+9x-8x+6=0

We add all the numbers together, and all the variables

-12x^2+6x+6=0

a = -12; b = 6; c = +6;
Δ = b2-4ac
Δ = 62-4·(-12)·6
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*-12}=\frac{-24}{-24}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*-12}=\frac{12}{-24}
=-1/2
$