Correct.

$\displaystyle{340}={20}\sqrt{{{t}+{273}}}\usepackage{cmbright}\Huge\$

$\displaystyle{17}=\sqrt{{{t}+{273}}}\usepackage{cmbright}\Huge\$

$\displaystyle{{17}}^{{2}}={{\left(\sqrt{{{t}+{273}}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{289}={t}+{273}\usepackage{cmbright}\Huge\$

$\displaystyle{t}={16}\usepackage{cmbright}\Huge\$ degrees Celsius