How do you solve by completing the square for $y=3x^2+24x+8$ ?

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Answer 1 · 2015-05-28T04:58:04

$y = 3(x^2 + 8x + 16) - 48 + 8$

$y = 3(x + 4)^2 - 40$

$(x + 4) ^2 = 40/3 -> (x + 4) = +- (2sqrt10)/(sqrt3)$ = $+- (2sqrt30)/3$

$x = -24/6 +- (2sqrt30)/18 = -6 +- (sqrt30)/9$

How do you solve by completing the square for y=3x^2+24x+8y=3x^2+24x+8?