Why the $-2-\sqrt[3]{-7}$ is a complex root??? When I used the Wolframalpha to  solve the equation $$x^3+6x^2+12x+1=0$$, the result was as below:

My question is "Why the $-2-\sqrt[3]{-7}$ is a complex root???"
 A: The equation is $(x+2)^3=7$, which is satisfied if $x+2\in\left\{\sqrt[\large3]{7},\sqrt[\large3]{7}\,e^{2\pi i/3},\sqrt[\large3]{7}\,e^{4\pi i/3}\right\}$.
$\sqrt[\large3]{7}$ is the real root and WolframAlpha gets this
$\sqrt[\large3]{7}\,\color{#C00}{e^{2\pi i/3}}$ which WolframAlpha gives as $\color{#C00}{(-1)^{2/3}}\sqrt[\large3]{7}$
$\sqrt[\large3]{7}\,\color{#C00}{e^{4\pi i/3}}=\color{#C00}{-(-1)^{1/3}}\sqrt[\large3]{7}$ and WolframAlpha apparently combines the $(-1)^{1/3}$ and $\sqrt[\large3]{7}$ into $\sqrt[\large3]{-7}$, which is confusing because the real cube root function is often defined so that $\sqrt[\large3]{-x}=-\sqrt[\large3]{x}$ .
A: Note that the equation $x^3=-7$ has three roots.
One of them is the real one and the other two are complex numbers.
subtracting those two complex roots from $-2$ result in two complex roots  of your equation $$x^3+6x^2+12x+1=0$$
A: $x^3+7=0$ can be rewritten as $x^3+(\sqrt[3]{7})^3=(x+\sqrt[3]{7})(x^2-\sqrt[3]{7}x+\sqrt[3]{49})=0$. Discriminant of the quadratic equation is $-3\sqrt[3]{49}$ so it has two complex roots. I do agree that Wolfram notation is confusing, they should specify which cube root it is as there are three. 
A: There are three cube roots of $-1$:  $-1, e^{i\pi/3},$ and $e^{-i\pi/3}$.
There are three cube roots of $-7$:  $-\sqrt[3]7, -\sqrt[3]7e^{i\pi/3}, $ and $-\sqrt[3]7e^{-i\pi/3}.$
So, to answer your question, as lulu indicated in comments, a cube root of $-7$ can be complex.
