x3+6x+5=0 allows us to find the roots.
This can be written x3+6x=-5, x(x2+6)=-5, x=-5/(x2+6).
Let x=0 on the RHS, then x=-⅚=-0.8333 approx.
Let x=-⅚ on the RHS, then x=-5/(25/36+6)=-180/241=-0.7469.
So we had 3 successive values of x: 0, -0.8333 and -0.7469.
It would appear then that there is a real root between 0 and -0.8333.
When x=0, x3+6x+5=5 and when x=-⅚, x3+6x+5=-125/216=-0.5787 approx.
The change of sign from 5 to -0.5787 means that the root is between 5 and -0.5787.
When x=-0.7469, x3+6x+5=0.1020...
So there is a change of sign between x=-⅚ (-0.8333) and x=-0.7469 (-180/241), corresponding to -0.5787 and 0.1020. This means the root lies between -0.8333 and -0.5787 or, more accurately, between -⅚ and -180/241. We bisect the interval between these to get another value for x, that is, we find the average of the two values: x=-2285/2892 (about -0.7901), giving x3+6x+5=-0.2339 approx.
So the root lies between x=-2285/2892 and -180/241, because we have a change of sign between -0.2339 and 0.1020. The process of bisection continues. The next value is x=-4445/5784 (-0.7685), which gives us x3+6x+5=-0.0649. The process continues indefinitely until we get the accuracy we need.
The formula x=-5/(x2+6) will also produce the real root if we apply it iteratively.
The root to 10 decimal places is x=-0.7601324178, when x3+6x+5=-1.72×10-12 (-0.00000000000172). This is very close to zero, which verifies the root as approximately correct.