Alan

Write this as 2sec²x - 2secx - 4 = 0

This factors as 2(secx + 1)(secx - 2) = 0

So secx = -1 and secx = 2

Given that sec = 1/cos,  can you take it from here?

Income to Tribune = $1.79x

Cost to Tribune     = $(0.74x + 982,000)

Profit = Income - Cost = $(1.05x - 982,000)

The following might help.

For a general quadratic of the form a*x² + b*x + c the maximum (or minimum) occurs at x = -b/(2a).  In this case that is -4/(-2) = 2.

Here's a slightly different approach, though it ultimately amounts to the same as Melody's:

With the assumption that both masses are equal and momentum is conserved the momentum equations can be written as:

Horizontal:   17 = v1*cos(45) + v2*cos(30)

Vertical:         0 = -v1*sin(45) + v2*sin(30)

These can be solved easily enough to find v1 ≈ 8.8 and v2 ≈ 12.4

However, this must be an inelastic collision, because kinetic energy is not conserved.  That is 1/2*17² ≠ 1/2*v1² + 1/2*v2².

2.55 is correct if you use 9.8 for g and zero for μ.  I used 9.81 for g and 0.1 for μ which is why our final numerical results are slightly different.  

If you use the kinetic energy approach you don't need the negative sign on the m*g*sin(30).  The sign in my approach indicates that the direction down the slope is negative, and that the acceleration up the slope is negative (i.e. it is actually a deceleration).

6x² - 7x - 3 = (3x + 1)(2x - 3)

Quite right Anonymous, there are 3600 seconds in a degree, not 360; brain fade on my part I'm afraid!  I've corrected my original post.

I guess I should note that this is the same as using kinetic energy = force*distance, but now you must use total force along the line of motion, not just the frictional force (in the case of the object on the horizontal floor the frictional force was the total force along the line of motion).