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????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
2019-03-28
源自:初識Python 7-4
正在回答
‘ / ’前面‘ ) ’少了
x1=(-b+math.sqrt(b*b-4*a*c))/(2*a)
x1
=
(
-
b
+
math.sqrt(b
*
4
a
c))
/
2
a)
x2=(-b-math.sqrt(b*b-4*a*c))/(2*a)????
x2
????
return?x1,x2
return
x1,x2
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2019-03-29
‘ / ’前面‘ ) ’少了
2019-04-01
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