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

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

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

We move all terms to the left:

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


Domain of the equation: (3x+5)!=0

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

3x!=-5

x!=-5/3

x!=-1+2/3

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


2x^2/(3x^2+5x)+(-9x-15)/(3x^2+5x)-4=0

We multiply all the terms by the denominator


2x^2+(-9x-15)-4*(3x^2+5x)=0

We multiply parentheses

2x^2-12x^2+(-9x-15)-20x=0

We get rid of parentheses

2x^2-12x^2-9x-20x-15=0

We add all the numbers together, and all the variables

-10x^2-29x-15=0

a = -10; b = -29; c = -15;
Δ = b2-4ac
Δ = -292-4·(-10)·(-15)
Δ = 241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{241}}{2*-10}=\frac{29-\sqrt{241}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{241}}{2*-10}=\frac{29+\sqrt{241}}{-20}
$