Решите пожалуйста уравнение)))))

Решите задачу:

$\frac{5}{2x+3} + \frac{3-2x}{x+2} =10 \\ \\ \frac{5(x+2)}{(2x+3)(x+2)} + \frac{(3-2x)(2x+3)}{(x+2)(2x+3)} - \frac{10(2x+3)(x+2)}{(2x+3)(x+2)} =0 \\ \\ \frac{5x+10}{2 x^{2} +7x+6} + \frac{9-4 x^{2} }{2 x^{2} +7x+6} - \frac{10(2 x^{2} +7x+6)}{2 x^{2} +7x+6} =0 \\ \\ \frac{5x+10+9-4 x^{2} -20 x^{2} -70x-60}{2 x^{2} +7x+6} =0\\\\5x+10+9-4 x^{2} -20 x^{2} -70x-60=0\\-24 x^{2} -65x-41=0/*(-1)\\24 x^{2} +65x+41=0\\D=4225-3936=289 \\ \sqrt{D} =17\\ x_{1} = \frac{-65+17}{24*2} = \frac{-48}{48} =-1\\\\$

$x_{2} = \frac{-65-17}{48} =- \frac{82}{48} =- \frac{41}{24} =-1 \frac{17}{24} \\ \\ 2 x^{2} +7x+6 \neq 0\\2 x^{2} +7x+6=0\\D=49-48=1\\ \sqrt{D} =1\\ x_{1} = \frac{-7+6}{4} =- \frac{1}{4} \\\\ x_{2} = \frac{-7-6}{4} =-3.25 \\ \\ \\ \\ \\ x=-1 \frac{17}{24} \\\\x=-1$