To find the base p bellow $$231_p \times 15_p = 4135_p$$
I did, $$(2p^2 + 3p + 1) \cdot (1p + 5) = 4p^3 + 1p^2 + 3p + 5$$ $$2p^2-12p-13=0$$ $$p = \{\frac{12 + \sqrt{248}}{4}, \frac{12 - \sqrt{248}}{4}\}$$
this looks weird to me, though, because I found two values for a base. So is it possible or did I do something wrong?