SOLUTION: What is the solution set for this equation? 4y(squared)-4y-15=0

Question 596078: What is the solution set for this equation?
4y(squared)-4y-15=0

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
What is the solution set for this equation?
4y(squared)-4y-15=0
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4y^2 - 4y - 15 = 0
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4y^2-10y+6y-15 = 0
2y(2y-5)+3(2y-5) = 0
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(2y-5)(2y+3) = 0
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y = 5/2 or y = -3/2
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Solution: {5/2 , -3/2)
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Cheers,
Stan H.
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