In the case of the square root of a negative real (say, the discriminant in a quadratic equation), there are considered both signs of the square root with 'i' (denoting the imaginary quantity) being outside the radical, i.e. $ x_{1,2}=\frac{-b\pm i\sqrt{\Delta}}{2a}$.
But for a complex quantity inside the square root, the negative sign is ignored, as it is stated that the complex quantity's square root will generate both the negative and positive quantities. So, for solving a quadratic equation with complex number as the discriminant, the roots are stated as $x_{1,2}=\frac{-b+\sqrt{\Delta}}{2a}$.
I am confused by the latter statement about the complex number, and also want a generalization for any root value, say cube root of a complex number, and so on.
Answer
What happens is the $x\in[0,+\infty)$, the symbol $\sqrt x$ means the only square root of $x$ which belongs to $[0,+\infty)$. So, if we want to talk about both square roots of $x$, we must mention $\sqrt x$ and $-\sqrt x$,
However, if $x\in\mathbb{C}\setminus[0,+\infty)$, there is no standard choice among the square roots of $x$. So, the symbol $\sqrt x$ may mean any square root of $x$.
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