Given:
I want to find the range of a projectile (ignoring wind resistance)
Hours of searching have given no useful results. Those that I thought were useful turned out to be incapable of accounting for negative angles. This formula in particular returns a range of 0 if the angle is 0, which simply doesn't make sense.
For example:
On
With horizental velocity $v_x=v_0\cos\theta$ and vertical velocity $v_y=-v_0\sin\theta-gt$ (here $\theta>0$) and integration we see $$x=\int_0^tv_0\cos\theta\,dt=v_0\cos\theta\,t\,\,\,,\,\,\,y=\int_0^t v_y\,dy=-v_0\sin\theta\,t-\frac12gt^2$$ Eliminating $t$ between them gives $$y=-\tan\theta\,x-\frac{g}{2v_0^2\cos^2\theta}x^2$$ the range $R$ is the point $(R,y_0<0)$ and you may continue from here!
Edit: The solution for $h=-y_0$ and $\theta<0$ after solving the equation $$\frac{g}{2v_0^2\cos^2\theta}d^2+\tan\theta\,d-h=0$$ is $$d=-\dfrac{v_0^2}{2g}\sin2\theta\left(1+\sqrt{1+\dfrac{2gh}{v_0^2\sin^2\theta}}\right)$$ which works for $\theta<0$ as you wanted.
On
You can try the following:
The initial horizontal velocity is $v\cos\theta$, the initial vertical velocity is $-v\sin\theta$ (note that it's positive). Before landing on the field, the coordinates of the rocket follows the following equations:
$$x=v\cos\theta\times t$$ $$y=h-v\sin\theta\times t-\frac{gt^2}{2}$$
The rocket lands at the moment $T$ such that $y(T)=0$. Solving the equation $y(X)=0$, one might be able to choose the root that "good" root for $T$. Put this into $x$ you get the range $x(T)$.
On
$V_y = 100 \sin(-12.5) = -21.6440$ m/s
$V_x = 100 \cos(-12.5) = 97.6296$ m/s
Solving for the time of flight t
$S = V_y\cdot t + \frac{1}{2}\cdot g\cdot t^2$
$-1.65 = -21.6440t + \frac{1}{2}\cdot -9.80665\cdot t^2$
$4.90333t^2 + 21.6440t - 1.65 = 0$
$t = \frac{-21.6440+\sqrt{21.6440^2-(4\cdot 4.90333\cdot -1.65})}{2\cdot 4.90333}$
$t = .07496$ seconds
Range is $x = V_x\cdot t$
$x = 97.6296\cdot .07496 = 7.3184$ m
In summary: $$x = V_x\cdot (\frac{V_y + \sqrt{V_y^2 - (4\cdot \frac{g}{2}\cdot S)}}{g})$$
Where:
$V_x = 100\cos(-12.5)$
$V_y = 100\sin(-12.5)$
$g = -9.80665$
$S = -1.65$
Assuming a projectile is launched at angle $\theta$ from horizontal from a height $y_0$ above the ground, where the ground is assumed to be represented by the line $y = 0.$ The measurement of the angle is such that $\theta = 0$ if the projectile initially is moving horizontally to the right, and otherwise the angle is measured in a counterclockwise direction from "horizontal to the right," so small positive angles are upward and small negative angles are downward.
The objective is to calculate the horizontal distance $d$ from the launch point to the point where the projectile hits the ground, that is, when the $y$ coordinate of the projectile reaches zero.
One way to compute $d$ is first to compute $t_f - t_i,$ the time that elapses from when the projectile is launched until the projectile hits the ground. We can arbitrarily set $t_i = 0.$ If the initial angle $\theta$ were positive, the projectile would first have to rise to the height where it stops rising, and then fall from that height to the ground. The upward velocity at time $t$ after the launch is $$v_y(t) = v_0 \sin\theta - gt.$$ If $t_0$ is the time such that $v_y(t_0) = 0,$ then $$t_0 = \frac{v_0 \sin\theta}{g}.$$ The height of the projectile at that time is $$y(t_0) = y_0 + \frac{v_0^2\sin^2\theta}{2g}.$$ The projectile then falls this far in the interval of time $t_f - t_0,$ that is, $$\frac12 g(t_f - t_0)^2 = y_0 + \frac{v_0^2\sin^2\theta}{2g},$$ and therefore \begin{align} t_f &= t_0 + \sqrt{\frac2g\left(y_0 + \frac{v_0^2\sin^2\theta}{2g}\right)} \\ &= \frac{v_0 \sin\theta}{g} + \sqrt{\frac{2y_0}{g} + \frac{v_0^2\sin^2\theta}{g^2}}. \end{align}
Arbitrarily setting the $x$ coordinate of the starting point to $0,$ the horizontal position of the projectile at any time $t$ is $$x(t) = v_0 t \cos\theta.$$ The desired distance $d$ is simply the $x$ coordinate at time $t_f$ (when the projectile hits the ground), so \begin{align} d &= v_0 t_f \cos\theta \\ &= v_0 \left( \frac{v_0 \sin\theta}{g} + \sqrt{\frac{2y_0}{g} + \frac{v_0^2\sin^2\theta}{g^2}} \right) \cos\theta \\ &= \frac{v_0 \cos\theta}{g} \left( v_0 \sin\theta + \sqrt{2gy_0 + v_0^2\sin^2\theta} \right). \tag1 \end{align}
In the case $y_0 = 0,$ the square root in $(1)$ is just $v_0 \sin\theta,$ and the formula can be simplified to the form $\frac{v_0^2 \sin(2\theta)}{g}$ which is often given as the range of a projectile.
In the case where $\sin\theta \neq 0,$ we can pull a factor of $v_0\sin\theta$ out of the expression in parentheses in $(1)$ to get $$ d = \frac{v_0^2\sin\theta\cos\theta}{g} \left( 1 + \sqrt{\frac{2y_0g}{v_0^2\sin^2\theta} + 1} \right). $$
Then, using the fact that $\sin\theta\cos\theta = \frac{\sin(2\theta)}{2},$ we get the formula you found on Wikipedia. But if $\theta = 0$ then $\sin\theta = 0$ and the Wikipedia formula has a division by zero, hence it does not evaluate to zero--it does not evaluate at all.