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solve y^2-4y=13 using completing the square method

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asked by Evans
  1. y^2-4y + (-4/2)^2=13+(-4/2)^2
    y-4y+4=13+4
    (y-2)^2-(sqrt(17))^2=0
    (y-2-sqrt17)(y-2+sqrt17)=0
    y= 2+sqrt17, y=2-sqrt17

    posted by bobpursley

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