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What is the solution set of the equation 2y^2 - 3y = 2 The solution set is {-1/2, [1]}
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this is a quadratic equation you can subtract 2 on both sides and put it in the form: ay^2+by+c=0
I'm still very confused. Would you be able to tell me exactly how to do it?
Are you asking me how to subtract 2 on both sides?
2y^2-3y-2=2-2 2y^2-3y-2=0
now you can try to factor or use quadratic formula
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