Same Lagrangian recipe as hard-margin, applied to the soft-margin primal. The dual comes out almost identical:

maximize Σᵢαᵢ − (1/2)ΣᵢΣⱼαᵢαⱼyᵢyⱼ(xᵢᵀxⱼ) subject to 0 ≤ αᵢ ≤ C, Σᵢαᵢyᵢ=0

The one change: αᵢ is now capped at C (not just ≥0). That cap is exactly where soft-margin's tolerance shows up in the dual.