What is the answer to this quadratic functions question?

h(t) = -16t² + v₀t + h₀

The v₀ is the initial velocity and h₀ is the initial height.

You are told the balls are served out of the machine at a height of 2 feet and an initial velocity of 130 ft/sec.

So now the equation becomes:

h(t) = -16t² + 130t + 2

You are asked to find the maximum height.  We can do this faster with calculus, but I'll show you the algebra way.

we want to put this into vertex form which has the form of:

h(t) = a(t - h)² + k

The vertex is the point (h, k).

The means the maximum height is "k" at the time "h".

To do that we need to do a complete the square step.  To prepare for that we need the right side to be in the form of (t² + bt).  So we can subtract 2 from both sides then divide -16 from both sides:

h(t) = -16t² + 130t + 2

h(t) - 2 = -16t² + 130t

-[h(t) - 2]/16 = t² - 8.125t

Now we complete the square using the following steps:

Start with t's coefficient (-8.125)

half it (4.0625)

square it (16.50390625)

Add that to both sides:

16.50390625 - [h(t) - 2]/16 = t² - 8.125t + 16.50390625

The right side can now be factored as a perfect square trinomial:

16.50390625 - [h(t) - 2]/16 = (t - 4.0625)²

Now solve for h(t) again, starting with subtracting the constant from both sides:

-[h(t) - 2]/16 = (t - 4.0625)² - 16.50390625

Multiply both sides by -16:

h(t) - 2 = -16(t - 4.0625)² + 264.0625

And finally, add 2 to both sides:

h(t) = -16(t - 4.0625)² + 266.0625

Rounded to 2DP, the maximum height is 266.06 feet.