求助，解方程的时候左右两边换个顺序解就变了是怎么回事呢，我把生成的两种解相减之后不等于0，方程是
Solve[x^3 + a*x^2 == 2, x]
Solve[2 == x^3 + a*x^2, x]

x1 = x /. sol1
x2 = x /. sol2
In[10]:= N[x1 /. a -> 3]
Out[10]= {0.732051 + 0. I, -2.73205 + 0. I, -1. + 0. I}In[11]:= N[x2 /. a -> 3]Out[11]= {-2.73205 + 0. I, 0.732051 + 0. I, -1. + 0. I}
In[14]:= N[x1 /. a -> -3]Out[14]= {3.19582, -0.0979117 + 0.785003 I, -0.0979117 - 0.785003 I}
In[15]:= N[x2 /. a -> -3]Out[15]= {-0.0979117 + 0.785003 I,
3.19582 - 4.44089*10^-16 I, -0.0979117 - 0.785003 I}In[18]:= N[x1 /. a -> 3.14]
Out[18]= {0.719831 + 7.40149*10^-17 I, -2.90262 +
1.11022*10^-16 I, -0.957215 - 2.22045*10^-16 I}In[19]:= N[x2 /. a -> 3.14]Out[19]= {-2.90262 + 0. I, 0.719831 + 0. I, -0.957215 + 0. I}
解的顺序和表示形式已经发生变化，两种形式下的解并不是一一顺序对应的。

Solve[x^3 + a*x^2 == 2, x]
{{x -> 1/3 (-a + a^2/(27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3) + (27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(1/3))}, {x -> -(
a/3) - ((1 + I Sqrt[3]) a^2)/(
6 (27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(1/3)) -
1/6 (1 - I Sqrt[3]) (27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3)}, {x -> -(a/3) - ((1 - I Sqrt[3]) a^2)/(
6 (27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(1/3)) -
1/6 (1 + I Sqrt[3]) (27 - a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3)}}
Solve[2 == x^3 + a*x^2, x]
{{x -> 1/3 (-a - a^2/(-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3) - (-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3))}, {x -> -(a/3) + ((1 + I Sqrt[3]) a^2)/(
6 (-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(1/3)) +
1/6 (1 - I Sqrt[3]) (-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3)}, {x -> -(a/3) + ((1 - I Sqrt[3]) a^2)/(
6 (-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(1/3)) +
1/6 (1 + I Sqrt[3]) (-27 + a^3 + 3 Sqrt[3] Sqrt[27 - 2 a^3])^(
1/3)}}

说不定结果就是一样的呢，你也可以手算下两组带符号的解化简后是不是一样的