A slightly longer comment although I have to admit that it misses the discussion here partially. Still I hope it provides some background to your question.

Generating equidistantly spaced grids is simply not always possible. The reason is that the absolute spacing of the possible floating point numbers depends on their magnitude [1]. The true, equidistant grid may not be representable in the precision that is used. Consequently, the spacing between consecutive points in a grid created with linspace can vary:

x = np.linspace(-50000.0, 50000.0, 1000001) # dx = 0.1
np.min(np.diff(x))
> 0.099999999991268851
np.max(np.diff(x))
> 0.10000000000582077

This does not necessarily mean that the calculation of the grid points is off by more than one ulp. It simply shows that the absolute value of one ulp changes over the extent of the grid.
It follows that even if the grid is calculated with higher precision (e.g. fma operations) or even arbitrary precision, we will hit a fundamental boundary: the finite precision floating point numbers closest to the "true" grid simply are not equidistantly spaced.
My estimate is that the grid spacing is at most off by ~N ulp. This should be far too small to have any consequences in calculations based on the grid points themselves.
However, if you write code that reconstructs the grid spacing from consecutive grid points, e.g. for scaling the value of an integral, you may notice it. I did when writing a unit test for Fourier transforms where I compared to analytical results. Take home message: use dx = (x[-1] - x[0])/(N-1) instead of dx = x[1] - x[0].

If you - for some reason - want the same grid spacing everywhere you may choose an appropriate new spacing. What generally seems to work (but not always) is one that fits the floating point spacing at the largest number in the grid. In other words add and subtract the largest number in your grid to the grid spacing:

dx = 0.1
N = 1000001
x0 = -50000.0

# "round" dx, assuming that the maximal magnitude of the grid is ~N*dx
dx = (dx + N * dx) - (N * dx)
> 0.10000000000582077
x = x0 + np.arange(N) * dx
np.max(np.diff(x)) - np.min(np.diff(x))
> 0.0

Curiosly, either by design or accident, arange() seems to do something similar as was mentioned by Eric. It creates a new grid spacing by adding and subtracting the starting point of the grid. This often has similar effect as adding and subtracting N*dx (e.g. if the grid is symmetric around 0.0). Consequently, arange() seems to trade keeping the grid spacing constant for a larger error in the grid size and consequently in the end point.

The above discussion somewhat ignored additional errors in the calculation of the grid points. If every grid point is the result of one multiplication and one addition, they should stay within one ulp (for x[i] != 0.0). If the grid is calculated by accumulation they may be larger.

If you are concerned about how arange() or linspace() calculates the grid, I would recommend to simply use

x = x0 + np.arange(N) * dx

which will yield an accurate representation of the grid you most probably want (within 1 ulp for x[i] != 0.0). If you do not have x0, N, and dx, try to infer them from other information on the grid.

To sum this all up: Even if your grid calculation is free of round-off error, the resulting grid using finite precision floating point numbers will only be exactly equidistant for appropriate grid spacing dx and starting values x0. On the other hand neither this effect nor round-off in the grid calculation should ever play a signification role in your calculations as the resulting errors are too small.

1. Comparison to calculations with decimal can be difficult as not all simple decimal step sizes are exactly representable as finite floating point numbers. E.g., in single precision 0.1000000014901161193847656250 is the number closest to 0.1. To make a good comparison you would have to compare the output of arange() to a grid calculated to infinite precision which has the single precision representation of 0.1 as grid spacing and not its decimal value.

2. Calculating the difference in single precision in your C code as below yields

float actual = -44.990000000000000;
float err1 = fabs(x1 - actual);
float err2 = fabs(x2 - actual);
\$ ./test_fma
x1 -44.9899978637695312  x2 -44.9900016784667969
err1 3.81470e-06  err2 0.00000e+00

If you calculate the ulp error you notice that the non-FMA solution is off by exactly one ulp, while the FMA solution yields the single precision float closest to the actual solution (error = 0 ulp). Losing one ulp should not matter for any application. If it does, use a data type with higher precision, e.g., double.

3. I wrote Python script that compares grids generated in different ways with a grid calculated with mpmath in high precision [2]. Maybe it is useful for further insights.

Cheers
Nils

PS: I realize that I went a little off-topic here and apologize for that. However, a while ago I spent some time trying to figure this issue out and thought that this may be an occasion to write some of it down.