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 

$$a^{ 2 }+{ a }^{ 2 }+b={ b }^{ 2 }+ab+b\\ { a }^{ 2 }-{ b }^{ 2 }=ab-{ a }^{ 2 }\\ \left( a-b \right) \left( a+b \right) =a\left( b-a \right) \\ \left( a-b \right) \left( a+b \right) +a\left( a-b \right) =0\\ \left( a-b \right) \left( 2a+b \right) =0\\ a\neq b,b=-2a\\ $$
