Find the displacement of mercury in a vertical tube
I assume the Hg is a competent barrier between the parcels of air and the parcels both have an initial pressure $P_0$ of 1 bar (760mm of Hg). Each air parcel has an identical initial energy content of $$ P_0V_0 = P_0A(0.45\mathrm{m}) $$

With hydrostatic equilibrium, $\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}}$ would all be due to the weight of the Hg, IOW, in mm of Hg terms, would be 100mm (0.1m) of Hg. -- This is your second equation. $$\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}} = 100\, \mathrm{\scriptstyle mmHg} $$From Boyle's Law we know that $$ P_{\mathrm{top}} = P_0 \frac{V_0}{V_{\mathrm{top}}} \quad\quad\text{and}\quad\quad P_{\mathrm{bot}} = P_0 \frac{V_0}{V_{\mathrm{bot}}} $$ Substituting, $$ \Delta P = P_0(\frac{V_0}{V_{\mathrm{bot}}} - \frac{V_0}{V_{\mathrm{top}}}) $$ or $$ (\Delta P) V_{\mathrm{top}} V_{\mathrm{bot}} = P_0V_0(V_{\mathrm{top}} - V_{\mathrm{bot}}) $$ Substituting again, $$ (\Delta P) A^2 ( 0.45\mathrm{m} - x )( 0.45\mathrm{m} + x ) = P_0A^2(0.45\mathrm{m}) ( (0.45\mathrm{m} - x) - (0.45\mathrm{m} + x) ) $$ eliminating $A^2$ and simplifying $$ \Delta P ( (0.45\mathrm{m})^2 - x^2) = -2P_0 x(0.45\mathrm{m}) $$ or in quadratic form $$ x^2 - 2\frac{P_0}{\Delta P}(0.45\mathrm{m})x - (0.45\mathrm{m})^2 = 0 $$ which would be $$ x^2 - (6.84\mathrm{m})x - 0.2025\mathrm{m}^2 = 0 $$ You can take it from there.