algebra precalculus - Connection between quadratic equations and complex numbers where $Delta > 0$ (of the quadratic formula)

I recently studied how the complex quadratic equation of $x^2-2x+5$ was solved:

$x^2-2x+5$">

Then I typed it in for the star of this question:

$$x^2-2x-5$$

And it used the quadratic formula, which I found boring. So I wondered: what will happen if I use the same method as the image above and just pretend it will have a complex solution. Will the imaginary numbers cancel out? What will happen?

So I did that, and my question is: how did this work so well? What is actually happening here? Is this a complex number thing, or just the fact that I'm splitting $x$ into two components?

Or did I just luck into the solution and actually have a subtle mistake?

Here are the steps I took.

$$x^2-2x-5=0$$

$$(a+bi)^2 - 2(a+bi) - 5 = 0$$

$$(a^2 + 2abi +b^2i^2) - 2a - 2bi - = 0$$

Reordering it to the 'template' of the image. (note: $b^2i^2 = -b^2$)

$$(a^2 - b^2 - 2a - 5) + i(-2b + 2ab) = 0$$

Alright, I'm on the right track as it is intuitive to me that only -5 should be the only changed value.

$$\begin{bmatrix} a^2 - b^2 - 2a - 5 = 0 \\ -2b + 2ab = 0 \end{bmatrix}$$

Here is where I venture of a little bit as my linear algebra is almost non-existent.

$$2ab - 2b = 0$$

$$2ab = 2b$$

$$2a = 2$$

$$a = 1$$

Since it's immediately visible what $b$ is in the same equation, I went to the other equation as I figured, it might have more information about $b$. We already have a believable value for $a$, so I subtitute it in there.

$$1^2 - 2*1 - b^2 - 5 = 0$$

$$-b^2 - 6 = 0$$

$$-6 = b^2$$

$$b = \sqrt{-6}$$

$$b = \sqrt{6}i \text{ or } b = -\sqrt{6}i$$

And substitute back for $x = a + bi$ we get:

$$x = 1 + (\pm\sqrt{6}i)i$$

$$x = 1 + (\pm\sqrt{6}i^2)$$

$$x = 1 + (\pm\sqrt{6}*-1)$$

$$x = 1 + (\pm\sqrt{6})$$

$$x = 1 \pm\sqrt{6}$$

Wait, that is correct (use the quadratic formula or something like SymPy to check, I use the website above called https://www.symbolab.com/solver/).

How is this possible? How is it possible that if I model this equation with complex numbers, I also get a valid solution? When isn't this possible?