A particle is moving along x-axis, At time t=0, its x-coordinate is $x=-4 \mathrm{m}$. Its velocity-time equation is $\mathrm{v}=8-2t$ where, $\mathrm{v}$ is in $\mathrm{m}/\mathrm{s}$ and t in seconds.
At how many times, particle is at a distance of $8 \mathrm{m}$ from the origin and find those times. 

Now, $8 \mathrm{m}$ distance from origin will be at two coordinates $x=8 \mathrm{m}$ and $x=-8 m$ From the diagram, we can see that particle will cross these two points three times, $t_1, t_2 and t_3$.

$t_1 and t_2$: At $x=8 \mathrm{m}$, displacement from the starting point is
$s=x_f-x_1=8-(-4)=12 \mathrm{m}$

Substituting in $s=ut+\frac{1}{2}a{t}^2$, we have
$12=8t-\frac{1}{2}\times 2\times t^2$

Solving this equation, we get
smaller time $t_1=2 \mathrm{s}$
and larger time $t_2=6 \mathrm{s}$
$t_3:$ At $x=-8 \mathrm{m}$, displacement from the starting point is
$s=x_f-x_i=-8-(-4)=-4 \mathrm{m}$
Substituting in $s=ut+\frac{1}{2}a{t}^2$, we have $-4=8t-\frac{1}{2}\times 2\times t^2$
Solving this equation, we get the positive time,
$t_3=8.47 \mathrm{s}$