Ответ:

6

Объяснение:

[tex](c - 3)(5c - 14) = c(c + 2) \\ 5 {c}^{2} - 14c - 15c + 42 = {c}^{2} + 2c \\ 5 {c}^{2} - {c}^{2} - 14c - 15c - 2c + 42 = 0 \\ 4 {c}^{2} - 31c + 42 = 0 \\ d = 31 {}^{2} - 4 \times 4 \times 42 = 961 - 672 = 289 \\ c1 = \frac{31 + \sqrt{289} }{2 \times 4} = \frac{31 + 17}{8} = \frac{48}{8} = 6 \\ c2 = \frac{31 - \sqrt{289} }{2 \times 4} = \frac{31 - 17}{8} = \frac{14}{8} = \frac{7}{4} = 1 \frac{3}{4} [/tex]