[tex]y = 2 {x}^{3} + 2.5 {x}^{2} - 6x + 2.001 \\ y' = 3\times 2{x}^{2} + 2 \times 2.5x - 6 = 6 {x}^{2} + 5x - 6 \\ 6 {x}^{2} + 5x - 6 = 0 \\ a = 6\\ b = 5 \\ c = - 6\\ D = {b}^{2} - 4ac = {5}^{2} - 4 \times 6 \times ( - 6) = 25 + 144 = 169 \\ x_{1} = \frac{ - 5 - 13}{2 \times 6} = - \frac{18}{12} = - \frac{3}{2} = - 1 \frac{1}{2} \\ x_{2} = \frac{ - 5 + 13}{2 \times 6} = \frac{8}{12} = \frac{2}{3} [/tex]

Ответ: 1)