10. $$y = 2x^3 - 3x^2 - 12x + 1$$
$$y' = 6x^2 - 6x - 12$$
Решим уравнение $$y' = 0$$:
$$6x^2 - 6x - 12 = 0$$
$$x^2 - x - 2 = 0$$
$$D = (-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9$$
$$x_1 = \frac{1 + \sqrt{9}}{2} = \frac{1 + 3}{2} = 2$$
$$x_2 = \frac{1 - \sqrt{9}}{2} = \frac{1 - 3}{2} = -1$$
Ответ: x = 2, x = -1