IF, invalid syntax error - syntax

Question

if volt.isalpha() or res.isalpha() or amp.isalpha():
What did I do wrong here? I get an INVALID SYNTAX, I am using this for a calculator program I am making. It calculates voltage, resistance, and amperage. But thats the easy part, I am just trying to make it fool proof. I have 3 variables in the code (volt, amp, res) that are inputted by the user. I just wanna make sure that they don't type in anything stupid. Like letters for e.g. ...
try:
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
print("")
print("You put a value for everything. You don't need the calculator.")
allowed = 0
if volt.isalpha() or res.isalpha() or amp.isalpha():
print("You typed in characters for one of the values, this calculator doesn't use letters.")
allowed = 0
def find_voltage(a,b): # V = I * R
voltage = a * b
return(voltage)`

Answer 1

You don't have an except block after try - it is required. Do something like:
try:
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
print("")
print("You put a value for everything. You don't need the calculator.")
allowed = 0
except ValueError:
print("Oops, you messed up.")
Additionally, the line
float(volt) >= 0 and float(res) >= 0 and float(amp) >= 0
doesn't do anything. You'll need to assign it to a variable, then check the results of the variable - if True, do one thing, if False, do something else.