Answer: 12 and \(\sqrt{24}\) (i.e. \(2\sqrt{6}\)), or 12 and \(-\sqrt{24}\)
Let the greater number be \(x\) and the smaller \(y\). Then \(y^2 = 2x\) and \(x^2 - y^2 = 120\).
\[x^2 - 2x - 120 = 0\]\[(x - 12)(x + 10) = 0\]If \(x = -10\), \(y^2 = -20\), which is impossible. So \(x = 12\) and \(y^2 = 24\), \(y = \pm\sqrt{24} = \pm 2\sqrt{6}\).
Check: \(144 - 24 = 120\) and \(24 = 2 \times 12\). The numbers are 12 and \(\sqrt{24}\) (or 12 and \(-\sqrt{24}\)).