Найдите корни уравнения: x^4-2x^2-8x-3=0

${x}^{4} - 2 {x}^{2} - 8x - 3 = 0 \\ {x}^{4} - 2 {x}^{3} + 2 {x}^{3} - {x}^{2} - 4 {x}^{2} + 3 {x}^{2} - 2x - 6x - 3 = 0 \\ {x}^{2} ( {x}^{2} - 2x - 1) + 2x( {x}^{2} - 2x - 1) + 3( {x}^{2} - 2x - 1) \\ ( {x}^{2} + 2x + 3)( {x}^{2} - 2x - 1) = 0 \\ {x}^{2} + 2x + 3 = 0 \\ {x}^{2} - 2x - 1 = 0 \\ d = {b}^{2} - 4ac = { (- 2)}^{2} - 4 \times 1 \times ( - 1) = 4 + 4 = 8 \\ x1 = \frac{2 + \sqrt{8} }{2} \\ x2 = \frac{2 - \sqrt{8} }{2} \\ x1 = \frac{2 + 2 \sqrt{2} }{2} = 1 + \sqrt{2} \\ x2 = \frac{2 - 2 \sqrt{2} }{2} = 1 - \sqrt{2}$
Ответ: х1 = 1+√2 ; х2 = 1-√2