简简单单反而对了。

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#1 Created at ... [Delete] [Delete and Lock User]

if(b*b-4*a*c)>0:
    x1=((-b+math.sqrt(b*b-4*a*c))/(2*a))
    x2=((-b-math.sqrt(b*b-4*a*c))/(2*a))
    return(x1,x2)
else:
    return("不对的参数")