Solve the following quadratic equation by factorization:
$3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5, x≠\frac{1}{3}, \frac{-3}{2}$

Given:

Given quadratic equation is $3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5, x≠\frac{1}{3}, \frac{-3}{2}$.

To do:

We have to solve the given quadratic equation.

Solution:

$3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5$

$\frac{3(3x-1)(3x-1)-2(2x+3)(2x+3)}{(2x+3)(3x-1)}=5$

$\frac{3(9x^2-3x-3x+1)-2(4x^2+6x+6x+9)}{6x^2-2x+9x-3}=5$

$\frac{27x^2-18x+3-8x^2-24x-18}{6x^2+7x-3}=5$

$19x^2-42x-15=5(6x^2+7x-3)$

$19x^2-42x-15=30x^2+35x-15$

$(30-19)x^2+(35+42)x-15+15=0$

$11x^2+77x=0$

$11(x^2+7x)=0$

$x^2+7x=0$

$x(x+7)=0$

$x=0$ or $x+7=0$

$x=0$ or $x=-7$

The values of $x$ are $0$ and $-7$.