Solve the following quadratic equation by factorization. 2xx−3+12x+3+3x+9(x−3)(2x+3)=0

Given
${\displaystyle \frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0}$
Step by step solution
${\displaystyle \Rightarrow \frac{2x(2x+3)+1(x-3)+3x+4}{(x-3)(2x+3)}}$
${\displaystyle \Rightarrow 4x^2+6x+x-3+3x+9=0}$
${\displaystyle \Rightarrow 4x^2+10x+6=0}$
${\displaystyle \Rightarrow 2(2x^2+5x+3)=0}$
${\displaystyle \Rightarrow 2x^2+5x+3=0}$
Complete Solution
${\displaystyle 2x^2+5x+3=0}$
${\displaystyle 2x^2+2x+3x+3=0}$
2x(x+1)+3(x+1)=0
(x+1)(2x+3)=0
${\displaystyle \Rightarrow x=-1}$
or
2x+3=0
${\displaystyle x=\frac{-3}{2}}$

Step 1
Given equation is ${\displaystyle \frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0}$
Upon simplification this can be written as:
${\displaystyle \frac{2x(2x+3)+1(x-3)+3x+9}{(x-3)(2x+3)}=0}$
Taking denominator to the right hand side
2x(2x+3)+1(x-3)+3x+9=0
Step 2
${\displaystyle \Rightarrow 4x^2+6x+x-3+3x+9=0}$
${\displaystyle \Rightarrow 4x^2+10x+6=0}$
${\displaystyle \Rightarrow 4x^2+4x+6x+6=0}$
${\displaystyle \Rightarrow 4x(x+1)+6(x+1)=0}$
${\displaystyle \Rightarrow (4x+6)(x+1)=0}$
${\displaystyle \Rightarrow 4x+6=0}$ or x+1=0
${\displaystyle \Rightarrow 4x=-6}$ or x=-1
${\displaystyle \Rightarrow x=\frac{-6}{4}}$ or x=-1
${\displaystyle \Rightarrow x=\frac{-3}{2}}$ or x=-1
Therefore x=-1.5 or -1

Step 1
Factorization method:-
$\begin{array}{}\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0\\ \Rightarrow \frac{2x(2x+3)+(x-3)+3x+9}{(x-3)(2x+3)}=0\\ \Rightarrow 4x^2+6x+x-3+3x+9=0\\ \Rightarrow 4x^2+10x+6=0\\ \Rightarrow 4x^2+4x+6x+6=0\\ \Rightarrow 4x(x+1)+6(x+1)=0\\ \Rightarrow (x+1)(4x+6)=0\\ \Rightarrow x+1=0or4x+6=0\\ \Rightarrow x=-1,-\frac{3}{2}\end{array}$
But $x\ne -\frac{3}{2}$
Step 2
Thus, x=-1 is the solution of the given equation