Restrictions for T in the equation y= (T+2)^2/5?

a^(2/5) is undefined when a is a negative number.
Therefore, a ≥ 0

In the equation y = (T + 2)^(2/5) :
(T + 2)^(2/5) must be defined, and thus (T + 2) ≥ 0

The restriction : T ≥ -2

T > -2

Adam is wrong. (T+2)^(2/5) is the same as saying "take the fifth root of (T+2) squared. You can take the fifth root of a negative number. A basic example: if T=-3, y = (-1)^(2/5). the fifth root of -1 is -1. -1 squared is 1, so y=1 (It is also valid to square first and then take the fifth root).

So T can be any real number.

In general, for a fractional exponent, as long as the fraction is in reduced form and the denominator is odd, it is valid for negative numbers. To realize why it must be in reduced form, think about x^(1/3) (aka the cube root). It is valid for negative x's. What about x^(2/6)? It's equal to x^(1/3), so it must also be valid for negative x's. That's why you must reduce.