Confused about a missing $\pm$ sign in texbook's answer (simple quadratic equation)

Question

The problem goes like this:

Working together, two cranes unload a barge in $t$ hours. What time does it take for each crane to unload the same barge on its own, provided that crane 1 spends $a$ less hours doing that?

My solution:

Let's say the barge's load is 1 unit of cargo.

The first crane's unloading speed is $V1$, the second's $V2$

When the two cranes work together, their unloading time is $t$:
$$\frac{1}{V1+V2}=t;$$ $$\frac{1}{t}=V1+V2$$

When they work separately, the first crane spends $a$ less hours unloading the barge:

$$\frac{1}{V1}=\frac{1}{V2}-a$$

Since our task is to find their standalone times, let's say $\frac{1}{V1}$ equals $x$, then $\frac{1}{V2}=x+a$. Then $$V1=\frac{1}{x}; V2=\frac{1}{x+a}$$

Then we get $$\frac{1}{t}=V1+V2;$$ $$\frac{1}{t}=\frac{1}{x}+\frac{1}{x+a}$$

This is easily transformed into a quadratic equation:

$$x^2+x(a-2t)-ta=0$$

The solution to which is $$x=\frac{-a+2t\pm\sqrt{a^2+4t^2}}{2}$$

This is the time the first crane spends to unload the barge on its own. To find the second crane's time, we only need to add another $a$, so to shorten the answer we could just put a plus-minus sign before $a$:

$$x=\frac{\pm{a}+2t\pm\sqrt{a^2+4t^2}}{2}$$

But the textbook's answer lacks the $\pm$ sign before the square root:

Why?

Answer 1

The other solution, where one of the cranes works at a negative rate, is physically reasonable but economically unintended.

The interpretation is that crane 2 is loading while crane 1 is reversing some of that work, by unloading.

Answer 2

Per Elliot G.'s helpful suggestion, we need to prove that

$$2t-a < \sqrt{a^2+4t^2}$$

Let's square both sides:

$$4t^2-4at+a^2 < a^2+4t^2;$$ $$-4at<0$$

Since we know from the problem's formulation that $a$ and $t$ are both positive, the inequality will always be true. This means that if we retain the $\pm$ sign in the answer, one of the two solutions will result in a "negative" time, which is unacceptable. Thus we can safely discard the minus part of the $\pm$ sign.