1. The problem statement, all variables and given/known data
A cannon having a muzzle speed of 1000 m/s is used to destroy a target on a mountaintop. The target is 2000m from the cannon horizontally and 800m above the ground. At what angle, relative to the ground, should the cannon be fired? Ignore air friction.

2. Relevant equations

3. The attempt at a solution
Xf=Xo+Vox t = Xo+Vo cosΘt
Yf=Yo+Voy t -1/2gt$$^2$$ = -Yo+VosinΘt-1/2gt$$^2$$

t=2000/1000cosΘ = 2/cosΘ

800=1000sinΘ(2/cosΘ) - 1/2g (2/cosΘ)$$^2$$
=2000tanΘ-1/2g 4/cos$$^2$$Θ
1/cos$$^2$$-sec$$^2$$=1+tan$$^2$$

800=tanΘ-2g(1+tan$$^2$$Θ)

ax$$^2$$+bx+c=0

Ok, what do I do to get the angle now, grr I'm drawing a blank, I don't have long to finish this :\.
 Set $$\tan{\theta}\equiv Z$$ You know have a quadratic equation in $$Z$$
 19.62z$$^2$$+2000z-780.38 or 20z$$^2$$+2000z-780 But now what?

 Quote by neutron star 19.62z$$^2$$+2000z-780.38 or 20z$$^2$$+2000z-780 But now what?
Don't round those off. That's a very bad idea, especially since you're going to deal with $$\tan{\theta}$$ soon and that function is sensitive to such small changes.

You have a quadratic equation, how do you solve a quadratic equation?

 Quote by RoyalCat Don't round those off. That's a very bad idea, especially since you're going to deal with $$\tan{\theta}$$ soon and that function is sensitive to such small changes. You have a quadratic equation, how do you solve a quadratic equation?
-b+or- sq root b^2-4ac all over 2a

I did that and got weird answers.
 I got x=-15.37 x=-3984.63

 Quote by neutron star I got x=-15.37 x=-3984.63
You plugged your numbers in wrong.

I got:

$$Z_1\approx 0.3887$$

This correlates to: $$\theta=21.241^o$$

$$Z_2\approx-102.3255$$

This correlates to: $$\theta=-89.44^o$$ which is utter nonsense.