Plz im really in trouble with thisfind the equation of another tangent to which is parallel to the tangent at (1,2)

I'm just going to take a dab at this.  I haven't done calculus in years.

So you found the equation of the tangent line for your equation at the point (1,2) which is:
y = 5 - 3x

Which is correct and all, but the question seems to be asking for "another tangent", so we are looking for another tangent line that is parallel to:
y = 5 - 3x

So we know that the equation of this line must also have a slope of "-3".

So if it must have a slope of "-3", then it must follow that:

-3 = -3 + 24x - 24x²

I got the above from setting our derivative equation equal to -3.

So if we solve the above equation we get:

24x - 24x² = 0
24x(1 - x) = 0

So "x" must be equal to 0 or 1. (These are the two points that have a slope of -3)
And since we already know about the point x = 1, we are interested in the point x = 0.  So from our original equation the point x = 0 has a y-value of 1, therefore we have the point (0,1).  Finally, we just need to find the equation of the line that has a slope of -3 and contains the point (0,1).

I'm sure you can do this part on your own.  The equation of that line is:

y = -3x +1
or
y = 1 - 3x (as you pointed out)

Whew, that was a nice refresher.  :)

I hope that was easy to follow!

Byou start by adding them all and setting them equal to 180 since that is half of the total degrees in a circle.(2x+8) + (3x-14) + (3x-14) = 1808x-20 = 1808x = 200x = 25