二次函数f(x)满足f(x+1)+f(x-1)=2x^2+4x,求f(x)

f(x)=ax²+bx+c
f(x+1)=a(x+1)²+b(x+1)+c
=ax²+2ax+a+bx+b+c
f(x-1)=a(x-1)²+b(x-1)+c
=ax²-2ax+a+bx-b+c
ax²+2ax+a+bx+b+c +ax²-2ax+a+bx-b+c
=2ax²+2a+2bx+2c
=2x^2+4x
2a=2
a=1
2b=4
b=2
2a+2c=0
c=-1
f(x)=x²+2x-1