If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$ 
If $a$ and$ b$ are two distinct real values such that 
  $$F (x)= x^2+ax+b $$
   And given that  $F(a)=F(b)$ ; find the value of $F(2)$


My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F(a)=F(b)$  implies   $2a^2=b^2+ab$, and thus
$F(2)= 4+2a+b $
What Now?  
Any help would be appreciated , thank you 

$F(2) = 2^2+2a+b = 4+2a+b$. From $2a^2 = b^2+ab \implies a^2-b^2 = ab - a^2 \implies (a-b)(a+b) = a(b-a) \implies (a-b)(2a+b) = 0\implies a = b$ or $2a+b = 0$. Since $a \neq b, 2a+b = 0 \implies F(2) = 4+0 = 4$. 
