fix typo in EscapedChars.md

Author: nikfrank

EscapedChars.md

@@ -15,7 +15,7 @@ in the ES2015 specification. The characters included in the list are the followi
 | `.`       | So that `new RegExp(RegExp.escape("."))` won't be matched against single characters like `"a"` but instead against an actual dot ("."), and more generally that `.` won't be treated as an "any character" control construct. |
 | `*`       | So that `new RegExp(RegExp.escape("*"))` won't throw a type error but instead match against an actual star ("*"), and more generally that `*` won't be treated as a "zero or more times" quantifier. |
 | `+`       | So that `new RegExp(RegExp.escape("+"))` won't throw a type error but instead match against an actual plus sign ("+"), and more generally that `+` won't be treated as a "one or more times" quantifier. |
-| `?`       | So that `new RegExp(RegExp.escape("?"))` won't throw a type error but instead match against an actual question mark sign ("?"), and more generally that `?` won't be treated as a "once or not at all" quantifier. Also that `new RegExp("("+RegExp.escape("?=")+")")` will match a literal question mark followed by an equals sign, instead of introducint a lookahead, and more generally that `?` won't make groups become assertions or non-capturing. |
+| `?`       | So that `new RegExp(RegExp.escape("?"))` won't throw a type error but instead match against an actual question mark sign ("?"), and more generally that `?` won't be treated as a "once or not at all" quantifier. Also that `new RegExp("("+RegExp.escape("?=")+")")` will match a literal question mark followed by an equals sign, instead of introducing a lookahead, and more generally that `?` won't make groups become assertions or non-capturing. |
 | `(`       | So that `new RegExp(RegExp.escape("("))` won't throw a type error but instead match against an actual opening parenthesis  ("("), and more generally that `(` won't be treated as a "start of a capturing group" logical operator. |
 | `)`       | So that `new RegExp(RegExp.escape(")"))` won't throw a type error but instead match against an actual closing parenthesis  (")"), and more generally that `)` won't be treated as a "end of a capturing group" logical operator. |
 | `[`       | So that `new RegExp(RegExp.escape("["))` won't throw a type error but instead match against an actual opening bracket ("["), and more generally that `[` won't be treated as a "start of a character class" construct. |