6 seconds after launch.
Step-by-step explanation:
You should make sure there aren't odd symbols in your question when you post it.
Assuming it is just the equation -5x^2 + 20x + 60, if you graph it you can see the path of the ball. at x=0 this is the starting point It then hits the ground when it touches the x axis. So this is finding zeroes of a quadratic.
You could do this several ways; find a way to factor it, guess and check, the quadratic formula or completing the square. I will complete the square as that is what I am most used to. If you would like to see another let me know.
The first step in completing the square is making sure the x^2 has a coefficient of 1, so now it has -5 so we have to factor out -5 from the equation.
-5x^2+20x+60 = -5(x^2 - 4x - 12)
Now you only really need to focus on the factored expression, but I will continue writing everything. The next step is to find (b/2)^2 where b is the coefficient of the x term. In this case -4. so (b/2)^2 = (-4/2)^2 = 4.
Again, focusing on x^2 - 4x - 12 you want to add and subtract (b/2)^2 from that expression. since you are adding and subtracting you are not changing the value, but we can use this.
-5(x^2 - 4x - 12) = -5(x^2 - 4x - 12 + 4 - 4)
Here you want to rearrange it a bit. Originally c was the constant term, -12, now you want it to be the positive (b/2)^2, which is positive 4. also you can combine the two other constant terms. the original c and -(b/2)^2, but I will hold off for now so I don't do too much at once.
-5(x^2 - 4x - 12 + 4 - 4) = -5(x^2 - 4x + 4 - 12 - 4)
Now you focus on x^2 - 4x + 4. Hope fully you recognize this is the same as (x-2)^2. This always happens at this step of the process. also notice -2 = b/2. Using only variables here are the first steps.
ax^2 + bx + c
a(x^2 + (b/a)x + (c/a))
a(x^2 + (b/a)x + (c/a) + (b/(2a))^2 - (b/(2a)^2)
a(x^2 + (b/a)x + (b/(2a))^2 + (c/a) - (b/(2a))^2)
a((x + b/(2a))^2 + (c/a) - (b/(2a))^2)
Again, being able to make x^2 - 4x + 4 into (x-2)^2 or x^2 + (b/a)x + (b/(2a))^2 into (x + b/(2a))^2 willa lways happen. this is because if you expand (x + b/(2a))^2 you always get x^2 + (b/a)x + (b/(2a))^2.
Now I would combine the -12-4
-5(x^2 - 4x + 4 - 12 - 4) = -5((x - 2)^2 - 12 - 4) = -5((x - 2)^2 - 16)
If you redistribute the -5 you get vertex form, but I am going to stop here because I will just undo that in the next step. So this is the form you want. Finally we can find when it equals 0. So, you set this equal to 0 and use algebra to solve.
-5((x - 2)^2 - 16) = 0
Divide both sides by -5
(x-2)^2 - 16 = 0
add 16 to both sides
(x-2)^2 = 16
Take the square root of both sides, but also count botht he positive and negative version of the answer. the reason is both 2 and -2 squared get you 4.
x - 2 = +/-4
I am using +/- to indicate I am using both positive and negative 4. Now though add 2 to both sides. since you have +/- 4 you are going to get two different results. -4+2 and 4+2
x = -2 and 6.
now, the question wants times after x=0 (the start) so you only get 6. so x=6, or in other words 6 seconds after launch.
Let me know if you have any questions.
