$-\sqrt{-3} * \sqrt{-3} = x$
$-(-3^{\frac{1}{2}}) * (-3^{\frac{1}{2}}) = x$
$ +3^{\frac{1}{2}} * -3^{\frac{1}{2}} = x$
$-3^-1 = x$, then $x = -3$
The correct result should be $3$ positive
And i can get that result with second development:
$-(\sqrt{3}i) * (\sqrt{3}i) = x$
$-\sqrt{3}i * \sqrt{3}i = x$
$-\sqrt{9}i^2 = x$
$-3 i^2 = x$, then $x = 3$
I want to know, what I am wrong with in the first way to solve it (using powers), I still do not have major knowledge of the school, therefore I need an elementary explanation, I have read some similar questions, but I have not understood the answers, in advance thanks .
$$-\sqrt{-3}=-(-3)^\frac{1}{2}\ne3^{\frac{1}{2}}$$ That would be the same as saying $$-i\sqrt{3}=\sqrt{3} \rightarrow i=-1$$ Which is obviously false.
It's not possible to solve your equation in real terms because: $$\sqrt{a}\sqrt{b}=\sqrt{ab}\iff a,b\ge0$$ And $a$ and $b$ are both $-3$ here, which clearly isn't $>0$.
Instead, like you did, use complex numbers to solve this