First a trivial algebraic simplification: $\begin{align}&{}\qquad (-2 x+2 α+1/(2 σ^2))\exp[(-(x-α)^2+(x-μ)/(2 σ^2))]\\ &{}\quad+(-2 x+2 β+1/(2σ^2))\exp[(-(x-β)^2+(x-μ)/(2 σ^2))] \\ \\ & = (-2 x+2 α+1/(2 σ^2))\exp[-(x-α)^2] \cdot \exp[(x-μ)/(2 σ^2)]\\ &{}\quad+(-2 x+2 β+1/(2σ^2))\exp[-(x-β)^2]\cdot\exp(x-μ)/(2 σ^2)], \end{align}$ so if this is $0$, then you can divide both sides by $\exp[(x-\mu)/(2\sigma^2)]$, since, as a value of the exponential function, that can never be $0$.
You have $ \left(-2(x-\alpha) + \frac{1}{2\sigma^2}\right) \exp(-(x-\alpha)^2) + \left(-2(x-\beta) + \frac{1}{2\sigma^2}\right) \exp(-(x-\beta)^2) = 0. $ A trivial substitution moves some complications into one place rather than two: $ \left(-2w + \frac{1}{2\sigma^2}\right) \exp(-w^2) + \left(-2(w-\gamma) + \frac{1}{2\sigma^2}\right) \exp(-(w-\gamma)^2) = 0. $ At this point I ponder whether some other trivial simplifications might help. E.g. you could cancel $\exp(-w^2)$ from both sides, and a few other things like that. But those don't seem potentially fruitful.
So, I'm thinking maybe Newton--Raphson or the like.
But: This looks like something that might have come from trying to find some MLEs. So are you sure you shouldn't be trying to solve for $\alpha$, $\beta$, and $\sigma$ instead?