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We get from your equation $$sqrt25-t^2=-5$$ since $$sqrt25-t^2geq 0$$ for $$25-t^2geq 0$$ so we get no solution.

Read carefully the first paragraph of what wikipedia has to say about square roots, noting the difference between the definitions of "square root" and "principal square root":

https://en.wikipedia.org/wiki/Square_root

$sqrtx $ doesn't mean "square root of x". $sqrtx $ means, by definition, $mathitthe principal square root $ of x, which means, $mathitthe square root of x with positive real part $ . Note that this definition can be applied to complex numbers also, e.g. for the following A-Level question:

if we did not require the principal square root to have a positive real part, then the answer would be: Solutions are 7-3i and -7+3i, whereas given the definition of principal square root, which is standard convention, we must throw away -7+3i, but keep 7-3i as a solution.

4 and −4 are square roots of 16.

However, $sqrt16 = 4$, $mathitnot pm 4$.

For further evidence in favour of my understanding of the conventional meaning of $sqrt $, In Rudin's PMA, Theorem 1.21 states:

For every real $x > 0$ and every integer $n > 0$ there is one and only one real y such that $y^n = x$. The number y is written $sqrt[n]x $ or $ x^1/n$.

Going back to the original questions,

"I know square root of real numbers cannot be negative. So t cannot be real."

I agree.

"But I don't know whether imaginary numbers' square root can be negative or not."

All complex numbers have two square roots (though they may be repeats). An imaginary number like $9i$ will be have two complex square roots. This is true for complex numbers also- they too have two complex square roots.

I actually think what you meant to ask here was, "can we square root a $mathitnegative$ number?" In the complex realm, yes, but we have to be careful if we are calculating the principal square root and are using the $sqrt $ sign. see:

https://en.wikipedia.org/wiki/Imaginary_unit#Proper_use

for why and when we have to be careful.

-4 has two complex square roots, namely $2i$ and $-2i$. Since the real part of both of these square roots is 0, either both or neither are considered principal square roots, depending on if you define the principal square root to be non-negative or strictly positive. To be honest I'm not sure, but if I had to guess, I would say they are both principal square roots of -4. I'm open to clarification on this point though.

If you ask if a number is negative, then in this context you are implicitly implying the number is real and negative, or integer and negative etc, because only real/integer/etc numbers can be negative. Complex numbers with imaginary part not equal to 0 cannot be negative in and of themselves, however, their $mathit real / imaginary parts $ can be negative.

"I think square roots can never be negative."

Square roots of real numbers can be negative. -4 is a square root of 16. But remember: $sqrtx$ means the $mathitprincipal$ square root of x, not just the square root of x. You would be correct to say, "the principal square root of a real number cannot be negative"

"Also, I don't think we can classify imaginary numbers as positive or negative."

Correct. But like I said you $mathitcan$ talk about their $mathit real / imaginary parts $

Let's begin with $sqrt25+t^2+5=0implies sqrt25+t^2=-5.$ If we square both sides, we get $25+t^2=25implies t^2=25-25=0implies t=0.$ Zero is real so $t$ is real.

Did someone read the original question? Since
$$sqrt25-t^2=-5Rightarrow 25-t^2=25iff t=0$$
and zero is no solution to the original equation, option (a) is the right answer.

All numbers have two square roots. By convention the positive square root is assumed to be the square root for positive reals. However in your example $t=0$ is O.K., since $sqrt25=pm 5$.

You can use the third binomial formula to see that $t=0$ is no solution.

$$sqrt25 - t^2 + 5 =0$$

$$sqrt25 - t^2 =- 5 $$

$$sqrt5 - tcdot sqrt5+t =- 5 $$

$$sqrt5cdot sqrt5 =- 5 $$

$$5 =- 5 qquad colorredtimes$$