Correct. You need to find t when $\displaystyle{16}=-{16}{{t}}^{{2}}+{52}{t}+{4}\usepackage{cmbright}\Huge\$ , or $\displaystyle{16}{{t}}^{{2}}-{52}{t}+{12}={0}\usepackage{cmbright}\Huge\$ .

Divide both sides by 4: $\displaystyle{4}{{t}}^{{2}}-{13}{t}+{3}={0}\usepackage{cmbright}\Huge\$

Use the Quadratic Formula: $\displaystyle{t}=\frac{{-{\left(-{13}\right)}\pm\sqrt{{{{\left(-{13}\right)}}^{{2}}-{4}\cdot{4}\cdot{3}}}}}{{{2}\cdot{4}}}\usepackage{cmbright}\Huge\$

This simplifies to: $\displaystyle{t}=\frac{{{13}\pm\sqrt{{121}}}}{{8}}=\frac{{{13}\pm{11}}}{{8}}={3}{\quad\text{or}\quad}\frac{{1}}{{4}}\usepackage{cmbright}\Huge\$

The ball is 16 feet high and going up when $\displaystyle{t}=\frac{{1}}{{4}}\usepackage{cmbright}\Huge\$ . It is coming down when $\displaystyle{t}={3}\usepackage{cmbright}\Huge\$ .