When solving the expression for displacement of an accelerating object and solving for the time variable: $$v_it+\frac{1}{2}at^2=d$$ $$v_it+\frac{1}{2}at^2-d=0$$ $$t=\frac{-v_i\pm \sqrt{v^2_i+2ad}}{a}$$ $$t=\frac{-v_i\pm v_f}{a}$$ where $a$ is acceleration, $t$ is time, $v_i$ is initial velocity, $v_f$ is final velocity, and $d$ is displacement.
How do I know if I should choose + or - on the last step? Is there a way do select the physical solution without any additional information?
The equation you are solving is $$v_i t + \frac{1}{2} a t^2 = d$$ Imagine what that means physically. You shoot a bullet up in the air, that reaches the highest point and then falls back to earth. The equation asks at what time it reaches a distance $d$ above you. It can reach that distance on the way up or on the way down, so the equation has (or can have) two solutions.
Now, in the real problem, you are not given $d$, but $v_f$. You can calculate $d$ from $v_f$, and then solve the equation. Or you can use the identity $v_f^2 = v_i^2 + 2ad$ as you did. But the sign of $v_f$ also tells you whether the bullet reached $d$ on the way up or on the way down. So $v_f$ gives information about which of the solutions is correct.
Of course, it is much easier to use the relationship:
$$ v_f = v_i + a t$$
to solve that for $t$ directly, without this detour.
