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判断+返回

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    #solve x

    if a != 0 and (b**2-4*a*c) >=0:

        x1 =(-b+math.sqrt(b**2-4*a*c))/(2*a)

        x2 =(-b-math.sqrt(b**2-4*a*c))/(2*a)

        return x1,x2


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