Pre Cal.

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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?

  • Pre Cal. -

    you have solved the "equation" but you were dealing with an inequation.

    the values of (-2 ± √10)/3 are the intercepts on a y-number line.

    so the solution would be
    y ≤ (-2 - √10)/3 OR y ≥ (-2 + √10)/3

    too bad it did not factor, it would have been easier to see.

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