Correct. Since $\displaystyle{1.5}=\pi\sqrt{{\frac{{L}}{{245}}}}\usepackage{cmbright}\Huge\$ , we know that $\displaystyle{{1.5}}^{{2}}={{\left(\pi\sqrt{{\frac{{L}}{{245}}}}\right)}}^{{2}}={\pi}^{{2}}{\left(\frac{{L}}{{245}}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{2.25}={\pi}^{{2}}{\left(\frac{{L}}{{245}}\right)}\usepackage{cmbright}\Huge\$ . This can be rewritten as $\displaystyle{2.25}={L}{\left(\frac{{\pi}^{{2}}}{{245}}\right)}\usepackage{cmbright}\Huge\$ , and then as $\displaystyle{2.25}{\left(\frac{{245}}{{\pi}^{{2}}}\right)}={L}\usepackage{cmbright}\Huge\$ .