Помогите решить неравенство с модулем, чем скорее тем лучше

Решите задачу:

$\left| \frac{x-7}{x^2+3x-10} \right | \geq 1\; \; \Rightarrow \; \; \left [ {{ \frac{x-7}{x^2+3x-10} \geq 1} \atop { \frac{x-7}{x^2+3x-10} \leq -1}} \right. \\\\1)\; \frac{x-7}{(x+5)(x-2)} -1 \geq 0\; ,\; \frac{x-7-x^2-3x+10}{(x+5)(x-2)} \geq 0\; ,\; \frac{-(x^2+2x-3)}{(x+5)(x-2)} \geq 0\\\\ \frac{x^2+2x-3}{(x+5)(x-2)} \leq 0\; ,\; \frac{(x-1)(x+3)}{(x+5)(x-2)} \leq 0\\\\Znaki:\; +++(-5)---[3]+++[1]---(2)+++\\\\x\in (-5,-3\, ]\cup [\, 1,2)\\\\2)\; \frac{x-7}{x^2+3x-10} +1 \leq 0\; ,$

$\frac{x-7+x^2+3x-10}{(x+5)(x-2)} \leq 0\; ,\; \frac{x^2+4x-17}{(x+5)(x-2)} \leq 0\\\\x^2+4x-17=0\\\\D=16+4\cdot 17=84=4\cdot 21\\\\x_1=\frac{-4-2\sqrt{21}}{2}=-2-\sqrt{21}\approx -6,58\; ;\; x_2=-2+\sqrt{21}\approx 2,58\\\\ \frac{(x+2+\sqrt{21})(x+2-\sqrt{21})}{(x+5)(x-2)} \leq 0\\\\+++[-2-\sqrt{21}]---(-5)+++(2)---[-2+\sqrt{21}]+++\\\\x\in [-2-\sqrt{21};-5)\cup (2;-2+\sqrt{21}\, ]$

$Reshenie:\; x\in [-2-\sqrt{21};-5)\cup (-5;-3]\cup [\, 1;2)\cup (2;-2+\sqrt{21}\, ]\\\\Celue\; resheniya:\; \; -6-4-3+1=-12$