Question 136159
Since you are interested in when the height is 40, set h to 40:
  h = -12t2 + 14t + 40
  40 = -12t2 + 14t + 40
Now solve this equation for t by
1) Getting one side equal of the equation to zero; and then
2) Using factoring or the quadratic formula to solve for t

To get one side equal to zero, subtract 40 from each side:
  40 - 40 = -12t2 + 14t + 40 - 40
which simplifies to
  0       = -12t2 + 14t + 0
or
  0       = -12t2 + 14t

Since the expression on the right is easily factored, factoring will be used in this solution. (The quadratic formula can also be used.)

Always start factoring with the greatest common factor (GCF). In this case the GCF is 2t.
  0       = 2t (-6t + 7)
This is all the factoring that can be done.

In order for this product (multiplication) to be zero, one of the factors must be zero. So
  2t = 0
or
  -6t + 7 = 0

Solve each of these equations for t. Divide both sides of
  2t = 0
by 2 and get
  2t/2 = 0/2
which simplifies to:
  t = 0
This says the ball was at height 40 at t = 0, the start (which we already knew).

For the other equation, subtract 7 from each side:
  -6t + 7 - 7 = 0 - 7
giving
  -6t + 0 = -7
or
  -6t = -7
Now divide both sides of the equation by -6:
  -6t/-6 = -7/-6
which simplifies to
  t = 7/6
or, writing 7/6 as a mixed number:
  t = 1 1/6
So the ball must return to a height of 40 after 1 1/6 seconds.