-x^3+6x^2-5=0 решите пж

$-x^{3} + 6x^{2} - 5 = 0\\-x^{3}+x^{2}+5x^{2}-5x+5x-5 =0\\-x^{2}(x-1) +5x(x-1) + 5(x-1) = 0\\-(x-1)(x^{2}-5-5) = 0\\-(x-1)=0\\x-1 = 0\\x=1$

$x^{2}-5x-5 = 0\\D = b^{2}-4ac=25+20 = 45\\\\x_{1} = \frac{-b+\sqrt{D}}{2a} = \frac{5+\sqrt{45}}{2} = \frac{5+3\sqrt{5}}{2}\\\\x_{2} = \frac{-b-\sqrt{D}}{2a} = \frac{5-\sqrt{45}}{2} = \frac{5-3\sqrt{5}}{2}$

Ответ: $x_{1} = 1; x_{2} = \frac{5+3\sqrt{5}}{2}; x_{3}=\frac{5-3\sqrt{5}}{2}$