*So hitting the ground means-- this literally means that h is equal to 0. Plus or minus the square root of negative 10 squared. The negative sign, negative times a negative, these are going to be positive. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. Divide the numerator and the denominator by 4, you get negative 5 over 4. So we need to figure out at which times does h equal 0. And so we have 8t squared minus 10t minus 25 is equal to 0. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. Now, we have to remember, we're trying to find a time. The roots of this equation -2 and -3 when added give -5 and when multiplied give 6. Problem 1: Solve for x: x 11x 7x 7 = 0 → 11x(x 1) 7(x 1) = 0 → (x 1)(11x 7) = 0 → x 1 = 0 or 11x 7 = 0 → x = -1 or x = -7/11.*

*So hitting the ground means-- this literally means that h is equal to 0. Plus or minus the square root of negative 10 squared. The negative sign, negative times a negative, these are going to be positive. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. Divide the numerator and the denominator by 4, you get negative 5 over 4. *So we need to figure out at which times does h equal 0. And so we have 8t squared minus 10t minus 25 is equal to 0. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. Now, we have to remember, we're trying to find a time. The roots of this equation -2 and -3 when added give -5 and when multiplied give 6. Problem 1: Solve for x: x 11x 7x 7 = 0 → 11x(x 1) 7(x 1) = 0 → (x 1)(11x 7) = 0 → x 1 = 0 or 11x 7 = 0 → x = -1 or x = -7/11.

And let's divide everything by negative 2, just so that we can get rid of this negative leading coefficient. So we only want to think about our positive answer. I wouldn't worry too much about the physics here.

So you divide the left hand side by negative 2, you still get a 0. And so this tells us that the only root that should work is 5/2. I think they really just want us to apply the quadratic formula to this modeling situation.

Substituting these values in the formula, x = [-2 ± √(4 – (4*1*-6))] / 2*1 → x = [-2 ± √(4 24)] / 2 → x = [-2 ± √28] / 2 When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square.

A ball is shot into the air from the edge of a building, 50 feet above the ground. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds.

There is, of course, one new skill that you must apply. Knowing how to take the square root of a number is essential to this lesson, so please review this lesson if needed. for one more example - Just to make sure that you really understand this concept before moving on.

## How To Solve Quadratic Formula Problems Research Paper Against Abortion

Let's look at our first example, which is an extremely basic equation. Our last example will demonstrate what happens when your last step does not result in a perfect square.

And I think in this problem they just want us to accept this formula, although we do derive formulas like this and show why it works for this type of problem in the Khan Academy physics playlist.

But for here, we'll just go with the flow on this example.

The physics, we go into a lot more depth and give you the conceptual understanding on our physics playlist.

But let's verify that we definitely are at a height of 0 at 5/2 seconds, or t is equal to 5/2.

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