√(10x^2-11x-6)^-1 помогите найти область определения

$y = \sqrt{\dfrac{1}{10x^{2}-11x-6}} \\ \\ $\left\{\begin{gathered} 10x^{2}-11x-6 \ne 0 \ (a) \\ \\ \sqrt{\dfrac{1}{10x^{2}-11x-6}} \ge 0 \ (b) \end{gathered} \right.$$

$(a): 10x^{2} - 11x - 6 \ne 0 \\ D = 121 + 240 = 361 \\ \\ x_{1} = \dfrac{11 + 19}{20} \ne \dfrac{3}{2} \ ; \ x_{2} \ne -\dfrac{2}{5} \\ \\ (b): \sqrt{\dfrac{1}{10x^{2}-11x-6}} \ge 0 \\ \\ \dfrac{1}{10x^{2}-11x-6} \ge 0 \\ \\ \dfrac{1}{(x-\dfrac{3}{2})(x+\dfrac{2}{5})} \ge 0 \\ \\ x \in (-\infty ; -\dfrac{2}{5})\cup (\dfrac{3}{2}; +\infty)$

$D(y) = (-\infty ; -\dfrac{2}{5})\cup (\dfrac{3}{2}; +\infty)$

Ответ: D(y) = (-∞; -2/5)∪(3/2; +∞)