Question

pls solve this equation

Answer 1

$2 \times ( {y }^{2} - 6 \times y) ^{2} - 8(y ^{2} - 6y + 3) - 40 = 0 \\ let \: ( {y}^{2} - 6y) \: be \: x \\ 2 \times {x}^{2} - 8x - 24 - 40 = 0 \\ 2 {x}^{2} - 8x - 64 = 0 \\ {x}^{2} - 4x - 32 = 0 \\ {x}^{2} - 8x + 4x - 32 = 0 \\ x(x - 8) + 4(x - 8) = 0 \\ (x - 8)(x + 4) = 0 \\ either \\ x - 8 = 0 \\ x = 8 \\ or \\ x + 4 = 0 \\ x = - 4$