| Author |
Comment/Response |
hj
|
 09/22/10 08:36am
If I get a list:
Solve[x^2 + 5 x - 3 == 0, x]
{{x -> 1/2 (-5 - Sqrt[37])}, {x -> 1/2 (-5 + Sqrt[37])}}
I can get one x with:
Solve[x^2 + 5 x - 3 == 0, x][[1]]
{x -> 1/2 (-5 - Sqrt[37])}
How do I get each result so I can put it in a function?
without the x -> ?
URL: , |
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