You forgot to take the absolute values of $\displaystyle{{h}}^{{5}}\usepackage{cmbright}\Huge\$ . Because $\displaystyle{{h}}^{{5}}\usepackage{cmbright}\Huge\$ has an odd exponent, it could have a positive or negative value. You need to write $\displaystyle\sqrt{{{{\left({{h}}^{{5}}\right)}}^{{2}}}}={\left|{{h}}^{{5}}\right|}\usepackage{cmbright}\Huge\$ because the square root symbol means take the positive root.