An office messenger on an errand covered 14 km using a bicycle. The speed of the bicycle was increased by 1 unit after covering the first 6 km. If the total journey was completed in 4 hours, find the speed for the remaining journey.

Let $v_1$ be the initial speed in km/h.
The total distance is $14 \text{ km}$.
The total time is $4 \text{ hours}$.

For the first part of the journey:
Distance $D_1 = 6 \text{ km}$.
Time $t_1 = \frac{D_1}{v_1} = \frac{6}{v_1}$.

For the remaining journey (second part):
Distance $D_2 = 14 \text{ km} - 6 \text{ km} = 8 \text{ km}$.
The speed was increased by $1 \text{ unit}$, so the new speed $v_2 = v_1 + 1 \text{ km/h}$.
Time $t_2 = \frac{D_2}{v_2} = \frac{8}{v_1 + 1}$.

The total time is the sum of the times for both parts:

$$T = t_1 + t_2$$ $$4 = \frac{6}{v_1} + \frac{8}{v_1 + 1}$$

Multiply the entire equation by $v_1(v_1 + 1)$ to eliminate the denominators:

$$4v_1(v_1 + 1) = 6(v_1 + 1) + 8v_1$$ $$4v_1^2 + 4v_1 = 6v_1 + 6 + 8v_1$$ $$4v_1^2 + 4v_1 = 14v_1 + 6$$

Rearrange into a standard quadratic equation form $ax^2 + bx + c = 0$:

$$4v_1^2 + 4v_1 - 14v_1 - 6 = 0$$ $$4v_1^2 - 10v_1 - 6 = 0$$

Divide by 2 to simplify:

$$2v_1^2 - 5v_1 - 3 = 0$$

The quadratic formula is $v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=2$, $b=-5$, $c=-3$.

$$v_1 = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$$ $$v_1 = \frac{5 \pm \sqrt{25 + 24}}{4}$$ $$v_1 = \frac{5 \pm \sqrt{49}}{4}$$ $$v_1 = \frac{5 \pm 7}{4}$$

Two possible solutions for $v_1$:

$$v_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3$$ $$v_1 = \frac{5 - 7}{4} = \frac{-2}{4} = -0.5$$

Since speed cannot be negative, we take $v_1 = 3 \text{ km/h}$.

The speed for the remaining journey is $v_2 = v_1 + 1$.

$$v_2 = 3 \text{ km/h} + 1 \text{ km/h}$$ $$v_2 = 4 \text{ km/h}$$

The speed for the remaining journey is $4 \text{ km/h}$.