If [tex]d_t=d_f[/tex], then
[tex]d_t=d_f+v_0t+\dfrac12at^2\implies0=v_0t+\dfrac12at^2[/tex]
We can factor out [tex]\frac12t[/tex] to write
[tex]\dfrac12t(2v_0+at)=0[/tex]
which means that either [tex]t=0[/tex] or [tex]2v_0+at=0[/tex]. In the latter case, we can directly solve for [tex]t[/tex] to get
[tex]2v_0+at=0\implies at=-2v_0\implies t=-\dfrac{2v_0}a[/tex]
