SOLUTION: How do I solve for r where: (1/(1+r))*(30/r)-100 = 0 I'm getting stuck on the steps in te middle..."the work"...

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Question 230820: How do I solve for r where:
(1/(1+r))*(30/r)-100 = 0
I'm getting stuck on the steps in te middle..."the work"...
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(1/(1+r))*(30/r)-100 = 0
(1/(1+r))*(30/r) = 100
Multiply thru by r*(1+r)
30 = 100r*(1+r)
30 = 100r^2 + 100r
10r^2 + 10r - 3 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 10x%5E2%2B10x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2810%29%5E2-4%2A10%2A-3=220.

Discriminant d=220 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-10%2B-sqrt%28+220+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2810%29%2Bsqrt%28+220+%29%29%2F2%5C10+=+0.241619848709566
x%5B2%5D+=+%28-%2810%29-sqrt%28+220+%29%29%2F2%5C10+=+-1.24161984870957

Quadratic expression 10x%5E2%2B10x%2B-3 can be factored:
10x%5E2%2B10x%2B-3+=+%28x-0.241619848709566%29%2A%28x--1.24161984870957%29
Again, the answer is: 0.241619848709566, -1.24161984870957. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+10%2Ax%5E2%2B10%2Ax%2B-3+%29

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Sub r for x

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How do I solve for r where:
(1/(1+r))*(30/r)-100 = 0
-----------------
Multiply thru by r(1+r) to get:
(30)-100r(r+1) = 0
---
-100r^2-100r+30 = 0
Divide thru by -10 to get:
10r^2 + 10r - 3 = 0
---
Use the quadratic formula:
r = [-10 +- sqrt(100 - 4*10*-3)]/20
---
r = [-10 +- sqrt(220)]/20
---
r = [-10 +- 2sqrt(55)]/20
x = [-1/2 + (1/10)sqrt(55)] or x = [-1/2 + (1/10)sqrt(55)]
========================================
Cheers,
Stan H.