Does the damping force oppose a spring's restorative force in damped oscillations?

If the damping force ($$F_d=-bv)$$ in damped simple harmonic motion opposes the spring's force $$(F_s=-kx)$$, why does the solution for the position as a function of time proceed from this equation?: $$-bv-kx=ma$$ ... And when substituting differentials with respect to $$t$$, we have the solution: $$x(t)=e^{\frac{-bt}{2m}}\cos(w^{'}t+\phi)$$ I'm wondering what the reason is for $$-bv-kx=ma$$ (I showed $$x(t)$$ for more context) when the damping force OPPOSES the spring.

The damping force doesn't oppose the restoring force, it opposes the motion (that's what $$-bv$$ means). So, when the mass is moving away from the center/equilibrium point the damping force and restoring force point in the same direction: towards the center. They switch to pointing in opposing directions when the mass reaches its furthest distance. Then the velocity changes direction, so the damping force now points away from the center, but the restoring force is always toward the center.
As for why the motions follows from that equation, that just your standard "sum up the forces and set them equal to m a." The damping force is $$-bv$$ the sping force is $$-kx$$, so the net force is $$-bv -kx$$.