Dropping a package with initial velocity

[Calpalned]

Homework Statement

A helicopter is ascending vertically with a speed of 5.30m/s . At a height of 100m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: v₀ for the package equals the speed of the helicopter.]

Homework Equations

1) v_f² = v₀² + 2aΔx
2) v_f = v₀ + at
3) x_f = x₀ + v₀t + ½(a)t²
4) Quadratic Formula

The Attempt at a Solution

Using the third equation with Δx = x_f - x₀ = -100, v₀ = 5.3 m/s and acceleration = -9.81, and the quadratic formula, I get the right answer of t = 5.09 seconds. However, I don't get why this other logical method does not work. Using equation one, I substitute v₀ with 5.3, a with -9.81 and Δx with -100. This leads to v_f² = 5.3² + 2(-9.8)(-100). v_f = -44.598 m/s. Plugging this value into the second equation I get -44.58 = 5.3 - 9.8t. t equals 4.09 seconds. What did I do wrong?

 

[Nathanael]

$\frac{44.58+5.3}{9.8}=5.09$ seconds

 

[Calpalned]

Thank you so much! I see: I subtracted incorrectly