Нужна помощь Желательно быстренько:)

Решите задачу:

$x = \frac{6}{y} \\ {( \frac{6}{y}) }^{2} + {y}^{2} = 20 \\ \frac{36}{ {y}^{2} } + {y}^{2} = 20 \\ 36 + {y}^{4} - 20 {y}^{2} = 0 \\ {y}^{4} - 20 {y}^{2} + 36 = 0 \\ {y}^{2} = a \\ {a}^{2} - 20a + 36 = 0 \\ a_{1} + a_{2} = 20 \\ a_{1} \times a_{2} =36 \\ \\ a_{1} =2 \\ a_{2} =18 \\ {y}^{2} = 2 \\ y =+- \sqrt{2} \\ x = \frac{6}{ +-\sqrt{2} } = \frac{6 \sqrt{2} }{2} = +-3 \sqrt{2} \\ \\ {y}^{2} = 18 \\ y =+- \sqrt{18} =+- 3 \sqrt{2} \\ x = \frac{6}{+-3 \sqrt{2} } = \frac{6 \sqrt{2} }{+-3 \times 2} = \sqrt{2} \\ \\ ответ: \\ (3\sqrt{2} ; \sqrt{2} ) \\ ( \sqrt{2} ;3 \sqrt{2} )$