Easy.
Equivalent negative capacitance at the terminals of this thing is -Ceq = C * (R2/R1).
However, GBW of this op-amp shall be well above (your frequency range) * ((R2+R1)/R1). For example, if R2=R1, and your range is 20MHz, I would be using an op-amp with 100MHz GBW, at least 50MHz. Another problem is that, for reliable operation when connected to any circuit, C shall be limited to the maximum capacity, which that op-amp is specified to be able to drive. The higher GBW of the op-amp, the smaller (usually) this Cmax is. R1 and R2 shall be relatively small, so that they do not detrimentally affect stability.
So, all in all, do not expect you can create a "huge" negative capacitance, but about --100pF is not that difficult. Of course it is easier if you need to compensate only a small positive capacitance with that emulated negative one.
And, of course, this op-amp amplifies the signal on its input terminals by (R2+R1)/R1 times, so input signal amplitude shall be limited so that the op-amp does is not clipping. And, needless to say, the same applies to the slew rate -- the op-amp shall not be slew rate crippled at your maximum (20MHz) frequency and maximum expected input amplitude.
