+6248 I don't see any really clever way of doing this. You can expand it using the binomial theorem to get
\(\left(2+\sqrt{3}\right)^9 = \sum \limits_{k=0}^9~\dbinom{9}{k}2^k~3^{(9-k)/2}\)
if you then expand all that out and combine things you'll have an expression that looks like
\(a + b \sqrt{3}\)
b above turns out to be 40545 so you'll need to take the square root of 3 out to 5 decimal places
