For details on L4 and L5, including a fascinating solution in which the equilateral triangle continuously changes size, but stays equilateral, please look at this answer.

L3 is one of the linear ones, first found by Euler, just like L1 and L2. It lies on the line between the two larger masses, but on the opposite side of the largest mass, so the line looks like this:

L3 - Primary - L1 - Secondary - L2.

Equations for L1, L2 and L3 look good on wikipedia; I don't know why the other linked answers only have 1 and 2 but not 3, since its formula isn't any more difficult than the others. I don't have time to proofread mathjax at the moment, but I will come back in a few hours and clean things up.

For details on L4 and L5, including a fascinating solution in which the equilateral triangle continuously changes size, but stays equilateral, please look at this answer. These and other variants of the triangle pattern work even when none of the masses is ignored. If the three masses are equal, the center of mass equals the geometric center. There are multiple definitions of the center of a triangle, but for an equilateral triangle, they all coincide. If the masses are not all equal, the center of mass will lie at the weighted average of the positions of the three bodies, and the whole triangle will spin around this point at one particular constant angular rate, $\omega^2 = G (m_1 + m_2 + m_3)/d^3$, where $d$ is the distance from the three masses to each other.

L3 is one of the linear ones, first found by Euler, just like L1 and L2. It lies on the line between the two larger masses, but on the opposite side of the largest mass, so the lineup looks like this:

L3 --- Primary --- L1 --- Secondary --- L2.

Typical equations for L1, L2, and L3, such as on Wikipedia, assume the smallest mass is effectively zero, but that turns out not to be necessary either. The three masses must lie on a line, and that line must rotate uniformly in a fixed plane, but we can find the three position solutions even when all three masses are significant. The equation to solve is a fifth-degree polynomial in the ratio of the distances between the masses. I won't derive anything, but just refer you to the book I read it in: Richard Battin's An Introduction to the Mathematics and Methods of Astrodynamics, chapter 8. Label the bodies not by their masses, but rather their positions ordered from left to right, on the $\xi$ axis, as $\xi_1 < \xi_2 < \xi_3$. Define $r_{ij}$ as the distance $\xi_j-\xi_i$ (guaranteed positive, from the way we labeled them), and let $\chi=r_{23}/r_{12}$. The value $\xi=0$ is the point about which the whole system rotates, $\xi_1$ must be negative, and $\xi_2$ may also be, depending on the exact ratios of the masses. After some algebra, one arrives at the equation $$(m_1 + m_2)\chi^5 + (3m_1 + 2m_2)\chi^4 + (3m_1 + m_2)\chi^3 - (m_2 + 3m_3)\chi^2 - (2m_2 + 3m_3)\chi - (m_2 + m_3) = 0$$ which has exactly one positive real root. Find that $\chi$ (and since this is a quintic, there is no neat, closed-form solution), then plug back into other expressions from the book (pages 366 to 369 of the 1999 edition) to find $\omega$, $\xi_1$, $r_{12}$, $\xi_2$ and $\xi_3$.