How do you complete the square to solve $4y^2 + 8y - 36 = 0$ ?

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Answer 1 · 2018-03-03T06:51:43

$color(red)(x = 2.1623, - 4.1623$

$4y^2 + 8y - 36= 0$

$4 (x^2 + 2x -9) = 0$

$x^2 + 2x = 9$

Add 1 to both sides.

$x^2 + 2x + 1 = 9 + 1 = 10$

$(x+1)^2 = (sqrt10)^2$

$x + 1 = +-sqrt10$

$x = +-sqrt10 - 1$

$color(red)(x = 2.1623, - 4.1623$

How do you complete the square to solve 4y^2 + 8y - 36 = 04y^2 + 8y - 36 = 0?