Let f(x)=ax2+bx+c,
then f(x-3)=a(x-3)2+b(x-3)+c=ax2-6ax+9a+bx-3b+c.
f(x)+f(x-3)=ax2+bx+c+ax2-6ax+9a+bx-3b+c=2x2+4x,
Therefore, 2ax2+x(2b-6a)+(2c+9a)=2x2+4x.
Matching coefficients:
x2: 2a=2, so a=1;
x: 2b-6a=4, 2b-6=4, 2b=10, b=5;
constant: 2c+9a=0, 2c+9=0, c=-9/2.
f(x)=x2+5x-9/2.
Check it out: f(x-3)=x2-6x+9+5x-15-9/2=x2-x+9/2.
Add x2+5x-9/2: 2x2+4x.
So f(x-3)+f(x)=2x2+4x; therefore f(x)=2x2+4x-f(x-3).