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?
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.