ПОМОГИТЕ ПОЖАЛУЙСТА. 2/x-3 - x/x-3 = 4x/x^2 - 9

$\frac{2}{x-3}$ - $\frac{x}{x-3}$ = $\frac{4x}{ x^{2} -9}$
$\frac{2-x}{x-3}$ - $\frac{4x}{ x^{2} -9}$ =0
$\frac{(x+3)(2-x)-4x}{(x-3)(x+3)}$ = $\frac{2x- x^{2} +6-3x-4x}{(x-3)(x+3)}$ = $\frac{-5x- x^{2} +6}{(x-3)(x+3)}$ = $\frac{- x^{2} -5x+6}{(x-3)(x+3)}$
По Th Виета: $\left \{ {{ x_{1} * x_{2} =c=6} \atop { x_{1}+ x_{2} =-b=5 }} \right.$
$\left \{ {{ x_{1} =2} \atop { x_{2}=3}} \right.$
$\frac{-(x-3)(x-2)}{(x-3)(x+3)}$ =- $\frac{x-2}{x+3}$