Х4 - 7Х2 +12 = 0 решить уравнение

Решите задачу:

${x}^{4} - 7 {x}^{2} + 12 = 0 \\ {x}^{2} = y \: \: \: ( \: zamena \: peremennoj)\\ {y}^{2} - 7y + 12 = 0 \: \: \: (kvadratnoe \: uravnenije) \\ d = {b}^{2} - 4ac = ( - 7) ^{2} - 4 + 12 = 49 - 48 = 1 \\ y1 = \frac{ - b + \sqrt{d} }{2a} = \frac{7 + \sqrt{1} }{2 \times 1} = \frac{7 + 1}{2} = \frac{8}{2} = 4 \\ y2 = \frac{ - b - \sqrt{d} }{2a} = \frac{7 - \sqrt{1} }{2 \times 1} = \frac{7 - 1}{2} = \frac{6}{2} = 3 \\ podstavim \: znacenie \: y \\ {x}^{2} = y1 \\ {x}^{2} = 4 \\ x = \sqrt{4} \\ x1 = 2 \\ x2 = - 2 \\ - - - - - - - \\ {x}^{2} = y2 \\ {x}^{2} = 3 \\ x1 = \sqrt{3} \\ x2 = - \sqrt{3}$