0 \ \ ; \ \ D = 1 + 48 = 49 \ ; \ x_{1} = 4, \ x_{2} = -3 \\ \\ (x+3)(x-4)>0 \\ x\in (-\infty ; -3)\cup (4;+\infty) \\ \\ \log_{2}(x^{2} - x -12) = \log_{2}2^{3} \\ \\ x^{2} - x -12 - 8 = 0 \\ \\ x^{2} - x - 20 = 0 \\ D = 1 + 80 = 81 \\ \\ x_{1} = \dfrac{1 + 9}{2} = 5 \ , \ x_{2} = -4 " alt=" \log_{2}(x^{2} - x -12) = 3 \\ \\ ODZ: x^{2} - x - 12 > 0 \ \ ; \ \ D = 1 + 48 = 49 \ ; \ x_{1} = 4, \ x_{2} = -3 \\ \\ (x+3)(x-4)>0 \\ x\in (-\infty ; -3)\cup (4;+\infty) \\ \\ \log_{2}(x^{2} - x -12) = \log_{2}2^{3} \\ \\ x^{2} - x -12 - 8 = 0 \\ \\ x^{2} - x - 20 = 0 \\ D = 1 + 80 = 81 \\ \\ x_{1} = \dfrac{1 + 9}{2} = 5 \ , \ x_{2} = -4 " align="absmiddle" class="latex-formula">
Ответ: x₁ = 5 , x₂ = -4