Произведение корней уравнения равно

$(x - 0.6)( {x}^{2} - 3x - 4) = (4x - 2.4) {(x + 1)}^{2} \\ (x - 0.6)( {x}^{2} - 3 x- 4) = 4(x - 0.6) ( {x}^{2} + 2x + 1) \\ (x - 0.6)( {x}^{2} - 3x - 4) - (x - 0.6)(4 {x}^{2} + 8x + 4) = 0 \\ (x - 0.6)( {x}^{2} - 3x - 4 - 4 {x}^{2} - 8x - 4) = 0 \\ (x - 0.6)( - 3 {x}^{2} - 11x - 8) = 0 \\ (x - 0.6)(3 {x}^{2} + 11x + 8) = 0 \\ x - 0.6 = 0 \: \: \: \: \: \: \: \: x = 0.6 \\ 3 {x}^{2} + 11x + 8 = 0 \\ d = {b}^{2} - 4ac = 121 - 4 \times 3 \times 8 = 25 \\ x1 = \frac{ - 11 + 5}{2 \times 3} = \frac{ - 6}{6} = - 1 \\ x2 = \frac{ - 11 - 5}{2 \times 3} = \frac{ - 16}{6} = - \frac{8}{3} \\ \\ 0.6 \times ( - 1) \times ( - \frac{8}{3} ) = 0.2 \times 8 = 1.6$
Ответ: 1,6