(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

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

5x/3x^2+9x/3x^2-2=0

We multiply all the terms by the denominator


5x+9x-2*3x^2=0

We add all the numbers together, and all the variables

14x-2*3x^2=0

Wy multiply elements

-6x^2+14x=0

a = -6; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-6)·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-6}=\frac{-28}{-12}
=2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-6}=\frac{0}{-12}
=0
$