The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?

To find where on the hill the canonball lands 
So 0.15x = 2 + 0.12x - 0.002x^2 
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0. 
So we have -0.002x^2 - 0.03x + 2 = 0  
I'll multiply through by -1 so we have
 0.002x^2 + 0.03x -2 = 0. 
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75

Yes i could be the length of the triangle