SOLUTION: Solve for R: 2(pi)R^2 + 2(pi)RH - S = 0 I have the answer in the textbook, but I can't seem to find how it was derived. I used the Quadratic Formula but the process got all c

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve for R: 2(pi)R^2 + 2(pi)RH - S = 0 I have the answer in the textbook, but I can't seem to find how it was derived. I used the Quadratic Formula but the process got all c      Log On


   

Question 1204132: Solve for R:
2(pi)R^2 + 2(pi)RH - S = 0
I have the answer in the textbook, but I can't seem to find how it was derived. I used the Quadratic Formula but the process got all convoluted and I kept getting the wrong answer.
Thank you!
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Solve for R+:

2piR%5E2+%2B+2piRH+-+S+=+0+

2piR%5E2+%2B+2piRH+=+S+

2pi%28R%5E2+%2B+RH%29+=+S+

R%5E2+%2B+RH=+S%2F%282pi%29+....complete square+R%5E2+%2B+RH+=>%28R%5E2+%2B+RH%2B%28H%2F2%29%5E2%29-%28H%2F2%29%5E2+=>%28R+%2B+H%2F2%29%5E2-%28H%2F2%29%5E2+

%28R+%2B+H%2F2%29%5E2-%28H%2F2%29%5E2=+S%2F%282pi%29+

%28R+%2B+H%2F2%29%5E2=+S%2F%282pi%29%2B%28H%2F2%29%5E2+

R+%2B+H%2F2=+sqrt%28S%2F%282pi%29%2B%28H%2F2%29%5E2%29+

R++=+sqrt%28S%2F%282pi%29%2B%28H%2F2%29%5E2%29-H%2F2+