F(x)=ax^2+bx+c
①F(x+2)=f(2-x)
→ a(x+2)^2+b(x+2)+c=a(2-x)^2+b(2-x)+c
→ 8a+2b=0
→ b=-4a；
②图象过点(0,3),
→ F(0)=3 → c=3；
③ax^2-4ax+3=0的解x1、x2,满足(x1)^2+(x2)^2=10
→ (x1+x2)^2-2*x1*x2=10
→ (4a/a)^2-2*(3/a)=10
→ a=1.
F(x)=x^2-4x+3.