Ta có: \(F\left( x \right) = a\,{x^2} + bx + c\)

Khi đó: \(F\left( 0 \right) = 2017 \Rightarrow a{.0^2} + b.0 + c = 2017\)\( \Rightarrow c = 2017\)

\(\begin{array}{l}F\left( 1 \right) = 2018 \Rightarrow a{.1^2} + b.1 + c = 2018\\ \Rightarrow a + b + 2017 = 2018\end{array}\) 

\( \Rightarrow a + b = 1 \Rightarrow a = 1 - b\,\,\,\,\left( 1 \right)\)

\(\begin{array}{l}F\left( { - 1} \right) = 2019\\ \Rightarrow a.{\left( { - 1} \right)^2} + b\left( { - 1} \right) + c = 2019\\ \Rightarrow a - b + 2017 = 2019\\ \Rightarrow a - b = 2\,\,\,\,\,\,\,\,\left( 2 \right)\end{array}\)

Thay \(\left( 1 \right)\) vào \(\left( 2 \right)\) , ta được: 

\(\begin{array}{l}\left( {1 - b} \right) - b = 2 \Rightarrow 1 - 2b = 2\\ \Rightarrow 2b =  - 1 \Rightarrow b = \dfrac{{ - 1}}{2}\end{array}\)

Thay \(b = \dfrac{{ - 1}}{2}\) vào \(\left( 1 \right)\) ta được: \(a = 1 - \left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{3}{2}\)

Khi đó: \(F\left( x \right) = \dfrac{3}{2}.{x^2} - \dfrac{1}{2}.x + 2017\)

\( \Rightarrow F\left( 2 \right) = \dfrac{3}{2}{.2^2} - \dfrac{1}{2}.2 + 2017\)\( = 6 - 1 + 2017 = 2022\)

Vậy \(F\left( 2 \right) = 2022.\)