A particle moves along the x-axis, its position at time t is given by

$x(t)= \frac{3t}{6+8t^2}$, $t≥0$,

where t is measured in seconds and x is in meters.

Find time at which acceleration equals 0.

I got the answer 0.5 and 0 but apparently i got the answer wrong. I know I need to use derivative twice. Can someone please help me

Calculation
$velocity = \frac {-3(4t^2-3)}{2(4t^2+3)^2}$

$Acceleration = 36x(16x^4+8x^2-3)$

answer 0.5 and 0 was found through wolfram