Решите уравнение 4x-6\x+2 - x\x+1= 9\x^2+3x+2

Решите задачу:

$\frac{4x-6}{x+2} - \frac{x}{x+1}= \frac{9}{ x^{2} +3x+2}\\\\ \frac{4x-6}{x+2} - \frac{x}{x+1} - \frac{9}{(x+2)(x+1)}=0\\\\ \frac{4 x^{2} +4x-6x-6- x^{2} -2x-9}{(x+2)(x+1)}=0\\\\ \frac{3 x^{2} -4x-15}{(x+2)(x+1)}=0\\\\x+2 \neq 0;x \neq -2 \\\\x+1 \neq 0;x \neq -1\\\\\\3 x^{2} -4x-15=0\\\\D=(-4) ^{2}-4*3*(-15)=16+180=196=14 ^{2}\\\\ x_{1} = \frac{4+14}{6}=3\\\\ x_{2} = \frac{4-14}{6}=-1 \frac{2}{3}$