Derive this someone? Not hard

Derive:

[f(a+h)-f(a-h)]/2h

From ax^2+bx+c, x1=a-h and x2=a+h

All steps pleas

f( a+h) = 2(a+h)^2 +b(a+h) +c
= 2(a^2 + 2ah +h^2) + ab + bh +c
= 2a^2 + 4a^2 h + 2h^2 + ab + bh +c
f(a-h) = 2(a-h)^2 +b(a-h) +c
= 2a^2 - 4a^2 h + 2h^2 + ab - bh +c
Did you mean [f(a+h)-f(a-h)]/ h //// I took out the 2 in the denominator
= [ 2a^2 + 4a^2 h + 2h^2 + ab + bh +c -( 2a^2 - 4a^2 h + 2h^2 + ab - bh +c) ] / h
= [ 2a^2 + 4a^2 h + 2h^2 + ab + bh +c - 2a^2 + 4a^2h -2h^2 -ab +bh -c ] / h
= [ 8a^2h +2bh ] / h
= 8a^2 + 2b