Why does this circuit give a quadratic equation for current? I'm trying to figure out the current passing through a circuit with a 120 V supply with 6 $\Omega$ wires and a 60 W bulb in series. After performing the following steps (taking P as the power of the bulb, $\varepsilon$ as the EMF, i as current, and r as the resistance of the wires):
$P = Vi$
$P = (\varepsilon - ir)i$
$P = \varepsilon i - i^2r$
$i^2r - \varepsilon i + P = 0$
Now substituting the values in:
$6i^2 - 120i + 60 = 0$
$i^2 - 20i + 10 = 0$
This gives
$i = 10+3\sqrt{10} \text{ and } i = 10-3\sqrt{10}$
both of which are real and positive values. So my question is, why exactly do we get two values of current?
 A: Here's how I would approach this.
First we work out the current through the lamp in the absence of resistive wires.
As $I=P/V$, we have $I=60/120 = 0.5A$
As the lamp draws $0.5A$, its resistance is $120/0.5=240\Omega$
If we now add the wires, the total resistance becomes $246\Omega$ and the current is$$i=120/246=0.488A$$
Note that this assumes 2 things: First, that the resistance of the lamp does not change when the current drops. That is certainly incorrect for an incandescent bulb. And secondly, that $120V$ is the voltage at which the lamp is assumed to use $60W.$
A: The reason is that the power of the bulb is given and no further information, eg working voltage, working current and working resistance, and so the way the problem is set up only current $\times$ voltage is a constant ie power depends on those two variables.  That being so without any further information both of your solutions are equally valid.
A: There is an insightful graphical way to see why you get two answers.
Let's assume that 60W is the power dissipated by your lamp in this particular circuit (this is what you did by using the corresponding equation in your system; in reality 60W would be the power dissipated at the nominal mains voltage of 120V, but we want to see "why" your solution gives two values.)
As stated in other answers, stating the power alone does not specify neither the voltage nor the current in your bulb. As a matter of fact there are infinite pairs of values V,I that can give a power of 60W.
We can plot them as a locus in the VI plane as the hyperbola V I = 60.
Now, the rest of the circuit your lamp is attached to has a characteristic, in the VI plane, that is linear and goes from the point V= E, I = 0 in open circuit to the point V = 0, I = E/r in short circuit.

The admissible points for the complete circuit are the intersections of the power hyperbola with this characteristic, and what you get in general are two points.
Note that by changing the values of the open circuit voltage or the resistance of the cables you can either shift or slant the generator's characteristic in such a way as to make it tangent to the hyperbola. In this case the two solutions becomes identical.
