The expression $2{\mathrm{x}}^2+4\mathrm{x}+7$ has minimum value m at $\mathrm{x}=\mathrm{\alpha }$. The ordered pair $\left(\mathrm{\alpha },\mathrm{m}\right)$ is

$\begin{array}{l}Given\exp ression2x^2+4x+7\\ comparingwitha{x}^2+bx+c\\ a=2,b=4,c=7\\ \therefore x=\frac{-b}{2a}=\frac{-4}{2\left(2\right)}=-1=\alpha \\ m=miimumvalue=\frac{4ac-b^2}{4a}\\ =\frac{4\left(2\right)\left(7\right)-16}{4\left(2\right)}\\ =\frac{56-16}{8}\Rightarrow \frac{-40}{8}\Rightarrow -5\\ \therefore (\alpha ,m)=(-1,-5)\end{array}$