From Förberedande kurs i matematik 1

If we complete the square of the expression, we have that

\displaystyle \begin{align} & x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\ & =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\ \end{align}

and because \displaystyle \left( x-\frac{5}{2} \right)^{2} is a quadratic, this term is at least equal to zero when \displaystyle x={5}/{2}\;. This shows that the polynomial's smallest value is \displaystyle \frac{3}{4}.