1. Feb 24, 2009

### rcrx

1. The problem statement, all variables and given/known data
A light bulb is wired in series with a 144Ohm resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 23.4W. What is ther resistance of the light bulb? Note that there are two possible answers.

2. Relevant equations
P = I^2R and I = V/R

3. The attempt at a solution
Solve for I to find Rbulb. I = (V / Rb + Rr)^2 x Rb.

Rearranged 23.4(Rb^2 + 288Rb + 144^2) = 120Rb

This is where I'm stuck, I need to rearrange that so I can get ax^2 + bx + c = 0

My math skills are a little weak since I have been out of school for a few years.

Thanks a lot!

2. Feb 24, 2009

### cristo

Staff Emeritus
Expand the brackets and subtract 120Rb from each side...

3. Feb 24, 2009

### rcrx

So, pardon me, but I'd get the following?

23.4Rb^2 + (23.4)288Rb + (23.4)144^2 - 120Rb = 0

4. Feb 25, 2009

### cristo

Staff Emeritus
Yes. Now, since you are dealing with numbers, you can simplify 23.4*288 and subtract 120 to obtain an equation in your required form.