How solder inductance affects RF circuits?
A few bits of information: -

A 10mm length of 1mm diameter wire has an inductance of about 6nH (check-out this calculator). A 1mm length of 0.1mm diameter wire has an inductance of about 0.6nH.

An 80MHz Colpitts LC oscillator I built recently used a fixed inductance of 56nH. It follows from this that the tuning capacitance was approximately: -

C = \$\dfrac{1}{4\pi^2\cdot f^2\cdot L}\$ = 71pF

This is near enough 90MHz (as per OP's question) to not worry too much.

If the inductance added by a solder connection is in the order of 1nH + 1nH (both ends of the inductor), the operating frequency would "de-tune" as the square root of the increase in inductance, therefore: -

Change in frequency is \$\sqrt{\dfrac{56}{58}}\$ = 0.983 - actual frequency would fall from 80MHz to about 78.6MHz.

This may seem a lot but if you are designing an oscillator to be accurate you would incorporate a phase-locked-loop (PLL) and varactor-diode tuning scheme to "mop-up" the factors you can't account for like inductor tolerance, leakage capacitance and heating effects changing operating frequency.

It's also unlikely you'd get an inductor off-the-shelf that had a tolerance that was within 5% so I would say that the inductance added by the connection (or even a short bit of circuit track) is nothing to worry about if you are designing an oscillator to be "spot-on" i.e. you would use a PLL/varactor-diode.

What about the effect of the solder introducing more capacitance to ground and causing a lowering of operating frequency? This could happen - a large mass of solder could make a noticable difference: -

C = \$\epsilon_0\epsilon_R\cdot\dfrac{A}{d}\$ where: -

\$\epsilon_0\epsilon_R\$ is about 8.854\$\times 10^{-12} \times 2\$ for normal PCB material and A = area of one plate and d = gap between plates. Given that the gap could be 0.5mm (to an inner layer) and a massive blob of solder might be 10 sq mm in area (simplification), the capacitance introduced will be in the order of about 0.4pF.

As a percentage of the 71pF (calculated above) this is smaller than the increase in inductance so I'm going to say it's trivial.

However, if you were operating ay 800MHz, the same tuned circuit would be 5.6pF and 7.1pF and this is a whole different story.