(2/x+1)+(3/5x-2)=0

Solution for (2/x+1)+(3/5x-2)=0 equation:

(2/x+1)+(3/5x-2)=0


Domain of the equation: x+1)!=0

x∈R



Domain of the equation: 5x-2)!=0

x∈R


We get rid of parentheses

2/x+3/5x+1-2=0

We calculate fractions


10x/5x^2+3x/5x^2+1-2=0

We add all the numbers together, and all the variables

10x/5x^2+3x/5x^2-1=0

We multiply all the terms by the denominator


10x+3x-1*5x^2=0

We add all the numbers together, and all the variables

13x-1*5x^2=0

Wy multiply elements

-5x^2+13x=0

a = -5; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-5)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-5}=\frac{-26}{-10}
=2+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-5}=\frac{0}{-10}
=0
$