What is the discriminant of `m^2 + m + 1 = 0` and what does that mean?

`m^2+m+1 = 0` is of the form `am^2+bm+c = 0`, with `a=1`, `b=1`, `c=1`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 1^2 - (4xx1xx1) = -3`

We can conclude that `m^2+m+1 = 0` has no real roots.

The roots of `m^2+m+1 = 0` are given by the quadratic formula:

`m = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)`

Notice that the discriminant is the part inside the square root. So if `Delta > 0` then the quadratic equation has two distinct real roots. If `Delta = 0` then it has one repeated real root. If `Delta < 0` then it has a pair of distinct complex roots.

In our case:

`m = (-b+-sqrt(Delta))/(2a) = (-1 +-sqrt(-3))/2 = (-1 +-i sqrt(3)) / 2`

The number `(-1+i sqrt(3))/2` is often denoted by the Greek letter `omega`.

It is the primitive cube root of `1` and is important when finding all roots of a general cubic equation.

Notice that `(m-1)(m^2+m+1) = m^3 - 1`

So `omega^3 = 1`

Given a quadratic equation (using `m` as the variable) in the form:
`color(white)("XXXX")``am^2 + bm +c = 0`

The solution (in terms of `m`) is given by the quadratic formula:
`color(white)("XXXX")``m = (-b+-sqrt(b^2-4ac))/(2a)`

The discriminant is the portion:
`color(white)("XXXX")``b^2-4ac`

If the discriminant is negative
`color(white)("XXXX")`there can be no real solutions
`color(white)("XXXX")`(since there is no real value which is the square root of a negative number).

For the given example
`color(white)("XXXX")``m^2+m+1 = 0`
the discriminant, `Delta` is
`color(white)("XXXX")``(1)^2 - 4(1)(1) = -3`
and therefore
`color(white)("XXXX")`there are no Real solutions to this quadratic.