Let after a time t, the cyclist overtake the bus. Then $96+\frac{1}{2} \times 2 \times t^2=20 \times t$ or $t^2-20 t+$ $96=0$

$\therefore t  =\frac{20 \pm \sqrt{400-4 \times 96}}{2 \times 1}$

$=\frac{20 \pm 4}{2}=8\,sec . \text { and } 12\,sec .$