Inductive reactance climbs with frequency and capacitive reactance falls. Put both in one circuit and there is exactly one frequency where they are equal — and being opposite in sign, they cancel completely. That frequency is resonance.

$f_{0} = \dfrac{1}{2\pi\sqrt{LC}}$, read aloud f-nought equals one over two pi root L C. $f_{0}$ is the resonant frequency in hertz, $L$ is the inductance in henries and $C$ the capacitance in farads. Note there is no resistance in it at all: R does not move where a circuit resonates. Take the root of the product first, then let 2π turn radians per second into hertz — drop that 2π and you have $\omega_{0}$, a real quantity in rad/s, but 6.28 times the number you were asked for.

Resistance decides how SHARPLY it resonates. $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ — Q equals one over R, root L over C. $Q$ is the quality factor, a bare number with no unit, $R$ is the loss resistance in the loop in ohms, and $L$ and $C$ are as before. The root $\sqrt{L/C}$ is the circuit's surge impedance in ohms, and dividing by R makes the whole thing dimensionless. More loss, lower Q, blunter peak — which is why a Q figure is really a statement about resistance.

And Q sets the width. $BW = \dfrac{f_{0}}{Q}$ — B W equals f-nought over Q, where $BW$ is the bandwidth in hertz between the two half-power points either side of the peak. A Q of 100 on a 1 MHz carrier gives a 10 kHz window; that is how a radio picks one station out of the air. Chain the three together — $f_{0}$, then $Q$, then $BW$ — and you have described a tuned circuit completely. It is also a rehearsal for the boss.