Помогите пожалуйста.

Решите задачу:

${x}^{2} + x - 6 = 0 \\ d = 1 - 4 \times 1 \times ( - 6) = 1 + 24 = 25 = {5}^{2} \\ x1 = \frac{ - 1 + 5}{1} = 4 \\ x2 = \frac{ - 1 - 5}{1} = - 6 \\ 10 {x}^{2} + 5x - 5 = 0 \\ d = 25 - 4 \times 10 \times ( - 5) = 25 + 200 = 225 = {15}^{2} \\ x1 = \frac{ - 5 + 15}{10} = 1 \\ x2 = \frac{ - 5 - 15}{10} = - 2 \\ 2 {x}^{2} - 12x + 18 = 0 \\ d = 144 - 4 \times 2 \times 18 = 144 - 144 = 0 \\ x1.2 = \frac{12 + - 0}{4} = 3 \\ {x}^{2} - 2x - 4 = 0 \\ d = 4 - 4 \times 1 \times ( - 4) = 4 + 16 = 20 \\ x1 = \frac{2 + \sqrt{20} }{2} \\ x2 = \frac{2 - \sqrt{20} }{2} \\ {x}^{2} + 2x + 5 = 0 \\ d = 4 - 4 \times 1 \times 5 = 4 - 20 = - 16net \: korney \\ 12 {x}^{2} - 12x \\ 12x \times (x - 1) = 0 \\ 12x = 0 \\ x1 = 0 \div 12 \\ x 1= 0 \\ x2 = x - 1 = 0 \\ x2 =1$