If in the last second a body freely falling without initial velocity has flown 3/4

If in the last second a body freely falling without initial velocity has flown 3/4 of the entire path, then the total time of the body falling is equal to …

Let’s write down the paths traversed by the body in time t and (t + 1 s):
S₁ = gt² / 2;
S₂ = g (t + 1) ² / 2;
Path in the last second:
S₂ – S₁ = g (t + 1) ² / 2 – gt² / 2 = (g / 2) * (t² + 2t + 1 – t²) = (g / 2) * (2t + 1);
The ratio of the last second path to the entire path:
(S₂ – S₁) / S₂ = (g (2t + 1) / 2) / (g (t + 1) ² / 2) = 3/4;
(2t + 1) / (t + 1) ² = 3/4;
4 (2t + 1) = 3 (t + 1) ²;
8 t + 4 = 3t² + 6t + 3;
3t² – 2t – 1 = 0;
t = (2+ root (4 + 12)) / 6 = 1 s.
The second root is negative and is not considered.
The total fall time is:
t + 1 c = 1 c + 1 c = 2 c.
Answer: 2.0 s