How can I get the numbers 2, 3, 7 and 8 to equal 22 using just normal
+, -, x and / maths functions (and brackets)?  All the numbers have to
be used, but they only can be used once.

one way i thought up was "27" - 8 + 3

other than that, i can't really think of a way...i can get 19,20,21,23,24...

also 23 + 7 - 8
but I don't think that qualifies. No, I'm stumped too

decimals...?

7 x 3 + .2 + .8 = 22

Well, it's just 2+3+7+8 if you do it in Base 9. . . .

Define A,B,C,D as integer variables; then combine it with the numbers
2,3,7,8 and obtain a set of integer equation problems:

2A + 3B + 7C + 8D - 22 = 0
subject to: A,B,C,D in Z
Solutions: (2,1,1,1), (3,3,1,0),... obtained with Excel Solver


Other possibilities: 
products (2*3), ...(7*8)
quocients (2/3), ... (7/8)

Example:
(2*3)A + 7B + 8C - 22 = 0
subject to: A,B,C in Z
Solutions: no solution in this case

well, if you follow the rules of math operations, (P,E,M,D,A,S)

2 + 3 * 8 - 4 = 22

3*8 = 24
2+24-4 = 22

eeehhg.... nvm..I used wrong numbers