Question

How do you solve `h = -16t^2 + 50t + 4` using the quadratic formula?

Answer 1

`h=(25+sqrt(689))/(16)`

`h=(25-sqrt(689))/(16)`

Explanation:

The quadratic formula is:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

where `ax^2+bx+c=0`

Here is your quadratic equation

`h=-16t^2+50t+4`

Looking at `ax^2+bx+c=0`
I can see that your values are...

`a=-16`
`b=50`
`c=4`

Now just put those values into the quadratic formula

`h=(-b+-sqrt(b^2-4ac))/(2a)`

`h=(-(50)+-sqrt((50)^2-4(-16)(4)))/(2(-16))`

`h=(-50+-sqrt(2500+256))/(-32)`

Negative divided by negative makes our numerator and denominator positive

`h=(50+-sqrt(2756))/(32)`

You might think we are finished here but remember to always check if you can simplify the square root. In this case `2756` can be divided by `4`, which will give us a `2` outside of the square root.

`h=(50+-sqrt(4*689))/(32)`

`h=(50+-2sqrt(689))/(32)`

Now notice how we can factor `2` out of our problem, thanks to simplifying the square root

`h=[2(25+-sqrt(689))]/[2(16)]`

Cancel the common factors

`h=[cancel2(25+-sqrt(689))]/[cancel2(16)]`

`h=(25+-sqrt(689))/(16)`

So our answers are...

`color(green)[h=(25+sqrt(689))/(16)]`

`color(green)[h=(25-sqrt(689))/(16)]`