Is $i$ equal to $\sqrt{-1}$?

The square root symbol does not make sense when applied to a negative number. Granted, people often write it that way for convenience. $i$ is defined by the equation $i^2=-1$. Yes, if there is a field containing a root of $x^2+1$ then if $i$ is such a root so is $-i$. When we define a field such as $\mathbb C$ we are implicitly choosing a root of $x^2+1$.

From a formal algebraic perspective, $\Bbb{C}$ is the field $\Bbb{R}[x]/(x^2+1)$ which is the field you get by adjoining to $\Bbb{R}$ the roots of $x^2 + 1$ and all the linear combinations of the form $a+ib$, where $a, b \in \Bbb{R}$. You really need to add just one of them (we call it $i$), because the other is just $0 + i(-1)$. Notice that it doesn't matter which we pick as $i$: if I pick one and you picked the other, we would never notice we made different choices.

Infact, the Galois group (automorphisms preserving the original field, thus fixing $\Bbb{R}$ in this case) of $\Bbb{C}$ has exactly two automorphisms: the identity, sending $i$ to itself, and the conjugation, sending $i$ to its opposite. By definition, automorphisms preserve structure, thus confirming the choice of $i$ doesn't really matter.

$\sqrt{\cdot}$ is seen in most cases as a function in $[0,\infty)$, so it have a unique value associated to each non-negative real. In this tradition we generally define $i:=\sqrt{-1}$, that can be thought informally as a kind of extension to the classical square root function.

Indeed we can extend the definition of the square root to any complex number, setting it as the principal value of $\sqrt z:=e^{1/2\ln z}$ with the same result.

However is true that $(-i)^2=-1$.