Помогите пожалуйста решить неравенство:

$\frac{156}{(5^{2- x^{2} }+1)^{2}} - \frac{136}{5^{2- x^{2} }+1} + 5 \geq 0 \\$
Замена $5^{2- x^{2} }+1 = t; t\ \textgreater \ 0 \\$
$\frac{156}{t^{2}} - \frac{136}{t} + 5 \geq 0 \\ \frac{156-136t+5t^{2} }{t^{2}} \geq 0 \\ 5t^{2} -136t + 156 \geq 0 \\$
$5t^{2} -136t + 156=0\\ \sqrt{D} = \sqrt{136^{2} - 4*5*156} = \sqrt{4^{2}*34^{2} - 4*5*4*39} = \\ = \sqrt{4^{2}(34^{2} - 5*39)} =4 \sqrt{34^{2} - 5*39} = 4 \sqrt{1156 - 195} = \\$
$= 4 \sqrt{961} = 4*31 = 124 \\$
$t_{1} = \frac{136 + 124}{10} = 26 \\ t_{2} = \frac{136 - 124}{10} = 1,2 \\$
t ∈ ( 0 ; 1,2 ] U [26 ; + оо)

$5^{2- x^{2} }+1 \leq 1,2$ или $5^{2- x^{2} }+1 \geq 26 \\$
$5^{2- x^{2} }\leq 0,2 \\ 5^{2- x^{2} }\leq 5^{-1} \\ 2- x^{2} \leq -1 \\$
$- x^{2} \leq -3 \\ x^{2} \geq 3 \\ x \leq -\sqrt{3} , x \geq \sqrt{3}$

$5^{2- x^{2} }+1 \geq 26 \\ 5^{2- x^{2} }\geq 25 \\ 5^{2- x^{2} }\geq 5^{2} \\$

$2- x^{2} \geq 2 \\ - x^{2} \geq 0 \\ x^{2} \leq 0 \\ x=0 \\$

Ответ: ( - оо; -√3 ] U { 0 } U [ √3 ; + оо )