Answer:
6 seconds
Explanation:
We know that projectile was already launched from the ground (0 ft) at a certain number of seconds. However, we want to find the other time at which the projectile returns to the ground. We can find this time by solving the quadratic and using the bigger root/solution.
Step 1: Factor -2x^2 + 6x + 36
The greatest common factor of -2, 6, and 36 is -2. Thus, we can factor it out. To do this, we divide every term by -2 and bring -2 outside a set of parentheses:
-2 / 2 = -1 = -x
6 / -2 = -3 = -3x
36 / -2 = -18
y = -2(x^2 - 3 - 18)
Step 2: Factor further by finding two numbers whose product equals a * c and adds up to b.
Let's focus on the x^2 - 3x - 18 for a moment. Currently, it's in standard form, whose general equation is:
ax^2 + bx + c., where
- a, b, and c are constants.
Thus, 1 is our a value, -3 is our b value, and -18
Two numbers whose product = a * c (aka 1 * -18) are 3 and -6. When you add 3 and -6, you get -3, which is the same as our b value (i.e., -3).
Thus, we've factored the quadratic and now have 0 = -2(x + 3)(x - 6).
Step 3: Divide both sides by -2 to clear -2:
(0 = -2(x + 3)(x - 6)) / -2
0 = (x + 3)(x - 6).
Step 4: Set terms equal to 0 to solve:
Our two terms are (x + 3) and (x - 6) and we can set both equal to 0 to solve for x, the time when the projectile is at 0 feet (once when it's launched and once when it returns to the ground):
Setting x + 3 equal to 0:
x + 3 = 0
x = -3
Thus, the projectile was on the ground at -3 seconds. Of course, this is impossible since you can't have negative time. Thus, we can rule this number out and we know the projectile must have reached the ground at the other time.
Setting (x - 6) equal to 0:
x - 6 = 0
x = 6
Thus, the projectile was also on the ground at 6 seconds. Because we can't use -3 (logically speaking), the projectile will hit the ground after 6 seconds.