∵f(x+1)=x²+4x-3
令x=0,则f(0+1)=0²+4*0-3=-3=f(-1)
∴f(-1)=-3
设x+1=t,则x=t-1
f(x+1)=f(t)=(t-1)²-4(t-1)+3
=t²-6t+8
∴f(x)=x²-6x+8