Question

find the root (x+1)²-16=0​

Answer 1

$(x + 1) {}^{2} - 16 = 0 \\ = > (x + 1) {}^{2} = 16 \\ = > x + 1 = \sqrt{16}$

=> x+1 = + 4

=> x = +4 -1

=> x = + 3

= x = +3 and -3

Hope it helps......

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Answer 2

$x { {}^{2} } + 2x + 2 - 16 = 0$

$x {}^{2} + 2x - 14 = 0$