Q) Solve the following quadratic equation by factorization method: $\frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)}=0, x\ne3,\frac{-3}{2}$

$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$ $2x(2x+3)+(x-3)+(3x+9)=0$ $4x^{2}+6x+x-3+3x+9=0$ $4x^{2}+10x+6=0$ $2x^{2}+5x+3=0$ $2x^{2}+2x+3x+3=0$ $2x(x+1)+3(x+1)=0$ $(2x+3)(x+1)=0$ Hence, x = -1, $\frac{-3}{2}$ Since, it is given that x $\neq \frac{-3}{2}$ Hence, x = - 1 is the only possible solution. Please do press "Heart" button if you liked the solution.