I'm stuck with this problem.

h = -16t^2 + 312
the "t" is seconds after object released and h stands for height. We're suppose to set the equation to 0
so far this is what I have:
-16t^2 + 312 = 0
-8(2t^2+39)= 0
-8=0 can be disqualified but how do I factor out the 2t^2 + 39 = 0 and find the answer? Please help

bzzt. Watch those signs. You meant to say

-8(2t^2-39)= 0
2t^2-39 = 0
2t^2 = 39
t^2 = 39/2
t = √(39/2)

Sorry about the signs. The answer was suppose to be 2 seconds. That's what I'm not understanding. I'm coming out with what you have- I moved over the 2 and that left the t^2 = 39/2 but that doesn't come out to be two seconds which is suppose to be the answer

well, surely you can see that

h(2) = -16(4) + 312 = 248
which is not zero. So, either 2 is wrong, or the equation is wrong.

h(t) = -16t^2 + 64
has h(2) = 0

thank you I'll check back with my teacher.

how do you solve x+y=2?

To solve the quadratic equation "2t^2 + 39 = 0," you can use factoring or the quadratic formula. Let's use factoring in this case.

To factor the equation, you need to find two numbers whose product is 39 (the constant term) and whose sum is 0 (since the coefficient of t^2 is already 1).

The possible factor pairs of 39 are:
1 and 39
-1 and -39
3 and 13
-3 and -13

Of these pairs, none of them sum up to 0, so it seems that factoring might not work in this case.

The next method is to use the quadratic formula, which can be used to find the solutions of any quadratic equation (ax^2 + bx + c = 0). The quadratic formula is:

t = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 2t^2 + 39 = 0, a = 2, b = 0, and c = 39. Plugging in these values into the quadratic formula, we get:

t = (0 ± √(0^2 - 4(2)(39))) / (2(2))

Simplifying further:

t = ±√(-4(2)(39)) / (4)

Now, let's calculate the discriminant (the part inside the square root):

√(-4(2)(39)) = √(-312) = √(-1) * √(312) = i√(312)

Therefore, the solutions are:

t = (±i√(312)) / 4

Now, simplifying further:

t = (±√(312)i) / 4
t = ±(√(4*78)i) / 4
t = (±2√78i) / 4
t = (±√78i) / 2

So, the solutions to the equation 2t^2 + 39 = 0 are t = (√78i) / 2 and t = (-√78i) / 2.