Question

y²-4-12=0 solve by competing square

Answer 1

${y}^{2} - 4 - 12 = 0 \\ \\ {y}^{2} - 16 = 0 \\ \\ {y}^{2} = 16 \\ \\ y = \sqrt{16} \\ \\ y = 4$







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Answer 2

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