Side comment first: yes, root of the polynomial is something that makes it to be equal to zero, so there is 0 in the right side of every equation.

**Square equations**

You're completely right up to the point where you're trying to find complex roots of a square equation. It is done using exactly the same way as for real roots:

` x = (-b ± √(b² - 4ac)) / 2a `

So if your equation is 1x² + 1x + 1 = 0, then a=1, b=1, c=1 and roots are

` (-b ± √(b² - 4ac)) / 2a = (-1 ± √(1 - 4)) / 2 = (-1 ± √(-3)) / 2`

Since √(-3) = i√3 (because i√3 x i√3 = (i x i) x (√3 x √3) = -1 x 3 = -3), the roots are -1/2 ± (√3/2)i

**It is possible to find complex roots your way**

You can find the roots the way you calculate things, but be more careful with imaginary part; let's represent x not as 'a + b' as you do but as 'a + bi', where both a and b are real numbers, and i is, well, square root of -1. Then we will get:

`1(a + bi)² + 1(a + bi) + 1 = 0 =>`

a² + 2abi + (bi)² + a + bi + 1 = 0 =>

a² + 2abi - b² + a + bi + 1 = 0 =>

`(a² - b² + a + 1) + (2ab + b)i = 0`

Now look at the number at the left side: it has real part (a² - b² + a + 1) and imaginary part (2ab + b)i. For this number to be zero, both parts should be zero (you cannot make zero out of real number by adding imaginary part to it).

So we got that

` a² - b² + a + 1 = 0 and (2ab + b)i = 0`

I'm going to look at the imaginary part first.

`(2ab + b)i = 0 =>`

2ab + b = 0 =>

`(2a + 1)b = 0`

Now we have two numbers (2a + 1 and b) such that their product is zero, so one of those numbers is zero. It cannot be b (try to substitute b = 0 -- the equation turns out to a² + a + 1 = 0, which has no real solutions, and a was a real number). Hence

` 2a + 1 = 0 => a = -1/2`

Now, when we know a, let's look at the real part.

`a² - b² + a + 1 = 0 =>`

1/4 - b² - 1/2 + 1 = 0 =>

b² = 3/4 =>

`b = ±√3/2`

We substituted x = a + bi, so we can get our two roots now: -1/2 ± (√3/2)i

**General situation**

There is a theorem that the number of (complex) roots a polynomial has is equal to its degree. Well, the whole thing is a bit less straightforward as some roots can repeat themselves, like in x² - 2x + 1 = 0 -- there is only one solution, x = 1, but since the same equation can be represented as (x - 1)(x - 1) = 0, the solution x = 1 is, well, counted twice -- once for each "x - 1". I'm not going deeper here, but what's important to know: if you take random polynomial of degree n, almost always it will have exactly n different complex roots.

It is not possible to calculate exact roots for general polynomial of degree over 4 -- not that we don't know how, it's proven theorem that the formula cannot exist.

**About my picture**

So, I have equations which look like ..x^24 + ..x^23 + ...and so on... + ..x^2 + ..x + .. = 0, where every '..' is either 1 or -1. I lose nothing assuming that x^24 always has 1 (think about it), so there are 24 coefficients left. It gives me 2^24 = 16777216 equations. I find approximate solutions for each of those, and each has exactly 24 solutions. In this way I'm getting 2^24 x 24 = 402653184 complex numbers. For each such number, if it is a + bi, I put a point with coordinates (a, b). And that's how I've got my picture!