A graph indicates that, for positive x, such values occur only between x=3.55 and x=3.6 and it appears linear between those extremes.

Evaluating the portion containing y = 125,

   3.57141                 124.99935

   3.57142                 124.9997

   3.57143                 125.00005

   3.57144                 125.0004

           

Since it seems to be linear rather than jumping, we can evaluate all the floors and divide the result into 125 to see what the outer, and therefore inner, x must be.

At x=3.572, the outermost floor is 35, so x= 125/35 ~= 3.57142857142857.

The function is certainly monotonically increasing for positive x, so this solution is unique in the first quadrant

For negative x a similar situation occurs on either side of x=-.3.2. Everywhere near that value the outermost floor evaluates to -39. Therefore the negative solution is x = 125/-39 ~= -3.20512820512821.