Posted by **Jennifer** on Tuesday, March 31, 2009 at 1:20pm.

Solve 3y^2 + 4y - 2 greater than/equal to 0

3y^2 + 4y - 2 ≥ 0

3y^2 + 4y - 2 = 0

y = [-4 ± sqrt (4^2 - 4(3)(-2)]/(2 • 3) = (-4 ± sqrt 40)/6 = (-4 ± 2 rad 10)/6 = (-2 ± rad 10)/3

Is this right?

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