Question

Помогите решить пожалуйста!
(-4x^2-x+5):(4x+5)

Answer 1

$\frac{( - {4x}^{2} - x + 5)} { (4x + 5)} \\ { - 4x}^{2} - x + 5 = 0 \\ {4x}^{2} + x - 5 = 0 \\ d = \sqrt{ {b}^{2} - 4ac } = \sqrt{1 - 4 \times 4 \times ( - 5)} = \sqrt{1 + 80} = \sqrt{81} = 9 \\ x1 = \frac{ - 1 + 9}{8} = 1 \\ x2 = \frac{ - 1 - 9}{8} = - 1.25 \\ 4x + 5 = 0 \\ 4x = - 5 \\ x = - 1.25 \\ \frac{ - {4x}^{2} - x + 5 }{4x + 5} = \frac{(x - 1)(x + 1.25)}{x + 1.25} = x - 1$