Question

resolver
2x^4-20x^2+32=0

Answer 1

$2 {x}^{4} - 20 {x}^{2} + 32 = 0 \\ 2 {t}^{2} - 20t + 32 = 0 \\ t = 2 \\ t = 8 \\ {x}^{2} = 2 \\ {x}^{2} = 8 \\ x = - \sqrt{2} \\ x = \sqrt{2} \\ x = - 2 \sqrt{2} \\ x = 2 \sqrt{2}$