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

1 Answer
ceenik
Apr 27, 2017

#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)]#

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