This equation is deducting from two equations we've learned 1. `v_f = v_i + at` 2. `∆x = v_it + 1/2 a t^2` From equations 2 we can rewrite as `1/2 a t^2+v_it - ∆x =0` Solve `t` from above quadratic equation: `a => 1/2a` `b => v_i` `c => -∆x` solve `t` `t = (-v_i±sqrt((v_i)^2-4(1/2a)(-∆x)))/(2(1/2 a))` `t = (-v_i±sqrt(v_i^2+2a∆x))/a` substitute `t` to equation 1: `v_f = v_i + a((-v_i±sqrt(v_i^2+2a∆x))/a)` `v_f = v_i + (-v_i±sqrt(v_i^2+2a∆x))` `v_f = ±sqrt(v_i^2+2a∆x)` square both side `v_f^2 = (±sqrt(v_i^2+2a∆x))^2`