Thank-you, your answer helps a lot.

My problem is the following: if a string has a left bracket, then keep everything before the bracket, else keep the whole string.
Example:
text1 text2 (text_to_discard) -> text1 text2
text1 text2 -> text1 text2

I could not find how to do this with regex only, but managed to do it doing the following:
Use a "If Then" converter
If (.*)\(.*
then $1
else $0

Explanation (for regex illiterate people like me): 
(.*) means any text, the brackets mean "group": if the pattern matches, then this group will be returned
\( means the left bracket (the back slash is necessary because "(" is a special character)
.*  means any text after the left bracket

In the If/then converter, if the pattern matches, then $1 refers to the (.*) part, the text before the left bracket.
If the text has no left bracket, then nothing matches, and $0 refers to the original string.

Maybe there is a way to do this just with a regex and not with the if / then converter. Please let me know.