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Hi Amy. I'm not sure why you say that these don't give a "regular x value": in fact you will get two x values corresponding to where the arrow is 120 yards up. One value (the smaller one) is where the arrow is still ascending and the other is where the arrow is descending. I would guess that the safest distance to launch your arrow from is the furthest from the wall in which the arrow can still cross over the wall. Rearrange your equation to make it equal zero:
And use the Quadratic Formula to find the two values of x:
I get a sensible answer. Give your calculator another attempt. Cheers, PS: A castle wall that is 120 yards high?!? Wow!  


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