How do i solve for r?

Question

#A = 2pir^2 +2pirh#

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Answer 1 · 2017-11-11T07:35:15

# r = -h/2 \pm (sqrt(pi(pih^2+2A))) / (2pi) #

#A=2pir^2+2pirh#

We want to put this into a form that looks like:

#ar^2+br+c=0#, where #a,b,c# are constants.

because we can then use the quadratic formula to solve it. So let's go back to the question:

#2pir^2+2pirh-A=0#

Notice that we can set:

Now let's use the quadratic formula:

# r = (-b \pm sqrt(b^2-4ac)) / (2a) #

Substituting in:

# r = (-(2pih) \pm sqrt((2pih)^2-4(2pi)(-A))) / (2(2pi)) #

# r = -h/2 \pm (sqrt(4pi^2h^2+8piA)) / (4pi) #

# r = -h/2 \pm (sqrt((4pi)(pih^2+2A))) / (4pi) #

# r = -h/2 \pm (sqrt(pi(pih^2+2A))) / (2pi) #