Physics question below?

Question

A light bulb is wired in series with a 124 ohm resistor, and they are connected across a 125.0 V source. The power delivered to the light bulb is 23.2 W. What are the two possible resistances of the light bulb?

a) lower value

b) higher value

Answer 1

P=V*I
V=I*R
so

P=I^2*R

Using Ohm's Law, and summing voltages around the circuit
124*I+R*I=125
or
I*(124+R)=125

also, the power in the bulb is
I^2*R=23.2
I^2=23.2/R
Squaring the first equation

I^2*(124+R)^2=125^2
(23.2/R)*(124+R)^2=15625
simplify

15625*R/23.2=
124^2+2*124*R+R^2

15376-425.5*R+R^2=0
solving the quadratic

R=385.6
or
R=39.87

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