In my AP Physics book, I came across a question which presents a projectile with an initial velocity of $v_0$, an initial launch angle of $\theta_0$, and a "random" horizontal line at height $h$. The question then asks to find the elapsed time between the projectile passing through height $h$ in both directions.
In both the book's solution and mine, we let $v_y=v\sin\theta$
For the book's solution, they substitute $v_{y0}$ into $\Delta x=vt+\frac12at^2$ where $\Delta x=h$, giving $\frac12at^2+v_{y0}t-h=0$. Solving for $t$ using the quadratic formula, $t=\frac{-v_{y0}\pm\sqrt{v_{y0}^2+2ah}}{a}$. Then, subtracting the two times to get the difference between the two results in the final answer of $$\Delta t=-2\frac{\sqrt{v_{y0}^2+2ah}}{a}$$ ($-2$ because $a$ is already negative, and we don't want a negative time)
For my solution, I start off with the fact that $v_h^2=v_{y0}^2+2a\Delta y$ where $v_h$ is the velocity at height $\Delta y$ which equals $h$. Then, because $a$ is constant, we can take $\Delta y=\frac12(v_{y0}+v_h)t$ and solve for $t$, giving us $t_1$. Finding the total time the projectile takes to complete its trajectory is $t_{total}=\frac{-2v_{y0}}{a}$ because by taking $v_y=v_{y0}+at$ and setting $v_y=0$, we get the time the projectile takes to reach the top of its arc. The final solution would then be given by $\Delta t=t_{total}-2t_1$, which, when unsimplified, is $$\Delta t=\frac{-2v_{y0}}{a}-\frac{4h}{v_{y0}+\sqrt{v_{y0}^2+2ah}}$$
Running both solutions through arbitrary numbers yield the same results, which makes me believe my answer is correct. However, I don't think an AP reader who would be grading this would immediately see this as correct, since it doesn't match what they're expecting. Checking my work, as well as answering if an AP reader would mark this as correct would be greatly appreciated.
If you rationalize the denominator of your second expression's second term and cancel/combine terms a bit, you'll get the first expression. See the details below. I don't know about AP physics readers, but certainly AP calculus readers would check this, since the second expression is obviously subject to rewriting by rationalizing the denominator. Also, AP readers will read along with your solution until you go wrong (if you did happen to go wrong), and then try to continue reading afterwards. That is, they'll deduct points (actually, avoiding adding points) for what's wrong, but then they'll see if what follows is OK given the error, unless the error greatly simplifies the subsequent work. They don't just look at an answer. FYI, I've been an AP calculus reader.
$$\frac{4h}{v_{y0}+\sqrt{v_{y0}^2+2ah}} \;\; = \;\; \frac{4h}{v_{y0}+\sqrt{v_{y0}^2+2ah}} \cdot \frac{{v_{y0}-\sqrt{v_{y0}^2+2ah}}}{{v_{y0}-\sqrt{v_{y0}^2+2ah}}}$$
$$ = \;\; \frac{4h\left( v_{y0}-\sqrt{v_{y0}^2+2ah} \right)}{v_{y0}^2 - v_{y0}^2 - 2ah} \;\;= \;\; \frac{2v_{y0} - 2\sqrt{v_{y0}^2+2ah}}{-a} $$
Therefore, your answer is equivalent to the book's answer:
$$\frac{-2v_{y0}}{a} \; - \;\frac{4h}{v_{y0}+\sqrt{v_{y0}^2+2ah}} \;\; = \;\; \frac{-2v_{y0}}{a} \; - \; \frac{2v_{y0} - 2\sqrt{v_{y0}^2+2ah}}{-a} $$
$$ = \;\; \frac{-2v_{y0}}{a} \; + \; \frac{2v_{y0} - 2\sqrt{v_{y0}^2+2ah}}{a} \;\; = \;\; \frac{-2v_{y0} + 2v_{y0} - 2\sqrt{v_{y0}^2+2ah}}{a} \;\; = \;\; -\;\frac{2\sqrt{v_{y0}^2+2ah}}{a} $$