(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Homework Help: Setting up Quadratic Equation

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