Помогите!!!как решать?

Решите задачу:

$\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{29}{10}\\\\t=\frac{x^2+1}{x},\; \; t+\frac{1}{t}-\frac{29}{10}=0,\; \; 10t+\frac{10}{t}-29=0\\\\10t^2-29t+10=0,\; t\ne 0\\\\D=441,\; t_1=\frac{29-21}{20}=\frac{2}{5},t_2=\frac{50}{20}=\frac{5}{2}\\\\a)\; \frac{x^2+1}{x}=\frac{2}{5},\; \; 5x^2+5=2x,\\\\5x^2-2x+5=0\\\\D=-96<0\; \to \; net\; kornej\\\\b)\; \frac{x^2+1}{x}=\frac{5}{2},\; \; 2x^2+2=5x$

$2x^2-5x+2=0\\\\D=9,\\\\x_1=\frac{5-3}{4}=\frac{1}{2},\; x_2=2\; \; -\; otvet$

$2)\; \frac{24}{x^2+2x-8}-\frac{15}{x^2+2x-3}=2,\; OOF:\; x\ne 2,x\ne -4,\; x\ne 1,\; x\ne -3\\\\t=x^2+2x-3,\; \; \to \; \; x^2+2x-8=t-5\\\\\frac{24}{t-5}-\frac{15}{t}=2,\; \; t\ne 0,\; t\ne 5\\\\24t-15(t-5)=2t(t-5)\\\\2t^2-10t-24t+15t-75=0\\\\2t^2-19t-75=0\\\\D=961,\; t_1=\frac{19-31}{4}=-3,\; t_2=\frac{50}{4}=\frac{25}{2}$

$a)x^2+2x-3=-3\\\\x(x+2)=0\\\\x_1=0,\; x_2=-2\\\\b)\; x^2+2x-3=\frac{25}{2}\\\\2x^2+4x-31=0\\\\D=264\\\\x_3=\frac{-4-2\sqrt{66}}{4}=-1-\frac{\sqrt{66}}{2}\\\\x_4=-1+\frac{\sqrt{66}}{2}$