Simplify the equation by dividing by 2:

Find the roots using the quadratic equation:

*Because the discriminant (part under the radical) comes to 0, we know our quadratic will have just one repeated solution.

Multiply both sides by x:

Put in standard form:

Factor quadratic:

Evaluate by first writing the quadratic equation.

The given equation is already in standard form.  Write the coefficients.

Simplify the radical using perfect squares.

The answer is:  

Write the quadratic equation in order to solve for the roots.

The coefficients of the parabola  will be used to substitute as values into the given formula.

Substitute the values into the equation.

Simplify the fraction.

Rewrite  using factors of perfect squares.

Replace the term.

The answer is:  

In order to solve this problem, we think about how to translate what is asked of us in the problem into a mathematical notion. The problem asks us to find when the ball returns to the ground. We know that the ball is launched from the ground at time , and if we plug  into our original function, we will find that at time ,  . Therefore, our "ground" is located at , which is also the horizontal axis. This means that if we are looking for when our ball returns to the ground, mathematically we are looking for values of  which make . In other words, we are looking for the zeros (or horizontal intercepts) of our function. 

To find the zeros, we first need to factor the function. The steps of this process can be seen below.

So, we can see that we have two factors -  and . Now, as usual when solving for zeros, we need to set the right side of our function equal to 0 and solve for . Because if either of these factors is zero, the entire function will equal zero (as anything multiplied by zero is zero), we can consider each factor separately when set to zero.

Thus, we can see that our two zeros are . Since we know that  is when the ball was first launched, we know that when the ball reaches the ground again must be . We can check this by plugging  back into the original function and ensuring that it evaluates to zero. 

Write the quadratic formula.

The coefficients of the variables can be determined by the given equation in standard form:  .

Substitute the terms into the equation.

Simplify the terms.

The answer is:  

Write the quadratic equation.

The given polynomial  is already in standard form.

Substitute the values into the quadratic formula.

Simplify the top and bottom.

The roots will be complex since we have a negative term inside the radical.

A possible root is:  

To find the vertex of the parabola, you have to get it into vertex form:

The vertex can then be found at the coordinate .

To get to vertex form, we have to complete the square.

Move the 7 over to the other side by subtracting 7 from both sides of the equation:

You're going to have to add something to both sides of the equation...

...the question now is what. What number, when put in the box, would create a "perfect square" on the right-hand side of the equation?

Well, a perfect square trinomial is one whose factors are the same, like so:

In other words, we're looking for .

Well, if is what goes in the box, and is just , then must equal . Now we can solve for .

And since goes in the box, we need to add 4 to both sides:

Now we can factor the right-hand side very neatly:

After we clean up a bit...

...we get:

That gives us a vertex of .

To complete the square, the equation must be in the form:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can factor the quadratic:

Take the square root of both sides:

Finish out the solution: