It appears that f(x) may be a quadratic, because 2(x^2+7) is quadratic. Assume f(x)=Ax^2+Bx+C, where A, B and C are constants to be determined.
f(x+2)=A(x+2)^2+Bx+2B+C and f(x-1)=A(x-1)^2+Bx-B+C.
f(x+2)+f(x-1)=2x^2+14=Ax^2+4Ax+4A+Bx+2B+C+Ax^2-2Ax+A+Bx-B+C.
Equating x^2 terms: 2=2A, so A=1;
equating x terms: 0=4A+B-2A+B; 2A+2B=0 so B=-A=-1;
equating constants: 14=4A+2B+C+A-B+C=5A+B+2C=5-1+2C, so 2C=10 and C=5.
Therefore, f(x)=x^2-x+5 substituting for A, B and C.
Check: f(x+2)=x^2+4x+4-x-2+5=x^2+3x+7; f(x-1)=x^2-2x+1-x+1+5=x^2-3x+7.
Add the two together: 2x^2+14=2(x^2+7). Checks out OK.