$\begin{align}  & 2{{x}^{2}}+3x+7=2\left[ {{x}^{2}}+\frac{3x}{2}+\frac{7}{2} \right] \\ & \text{Completing the square for }{{x}^{2}}+\tfrac{3x}{2}+\tfrac{7}{2} \\ & =2\left[ {{x}^{2}}+\tfrac{3x}{2}+{{\left( \tfrac{3}{4} \right)}^{2}}+\tfrac{7}{2}-{{\left( \tfrac{3}{4} \right)}^{2}} \right] \\ & =2\left[ {{\left( x+\tfrac{3}{4} \right)}^{2}}+\tfrac{47}{16} \right] \\ & \text{As }x\text{ is real, the term }{{\left( x+\tfrac{3}{4} \right)}^{2}}\text{ is always positive } \\ & \text{and hence the functionis positive} \\ & \text{for all real value of }x.\text{ } \\ & \text{The minimum value is }2\left( \tfrac{47}{16} \right)=\tfrac{47}{16} \\\end{align}$