$$\sqrt {{{\left( {0.798} \right)}^2} + 0.404 \times 0.798 + {{\left( {0.202} \right)}^2}} $$       $$ + 1$$ $$ = ?$$

According to question,
$$\sqrt {{{\left( {0.798} \right)}^2} + 0.404 \times 0.798 + {{\left( {0.202} \right)}^2}} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798} \right)}^2} + {{\left( {0.202} \right)}^2} + 2 \times 0.202 \times 0.798\,} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798 + 0.202} \right)}^2}} + 1\,\,\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab} \right]$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( 1 \right)}^2}} + 1 \cr & \Rightarrow 1 + 1 \cr & = 2 \cr} $$