import math #2元1次方程解法 #判断 b^2-4ac;解法 (-b[+-]math.sqrt(b^2-4ac))/2a def quadratic(a, b, c): tmp = b**2-4*a*c if tmp < 0 : return None if a==0 : return False return (-b+math.sqrt(tmp))/(2*a),(-b-math.sqrt(tmp))/(2*a) print(quadratic(2,3,1)) print(quadratic(1,3,-4))
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