Express $x^2+5x+7$ in the form $(x+a)^2+b$. Answer ... Hence, explain why the expression will never be negative. Answer ...

1. Use the square form that matches the $x^2+5x$ terms.

Half of $5$ is $\dfrac{5}{2}$, so
$(x+\dfrac{5}{2})^2=x^2+5x+\dfrac{25}{4}.$

2. Adjust the constant term to keep the expression equal to the original.

$x^2+5x+7=(x+\dfrac{5}{2})^2-\dfrac{25}{4}+7$
$=(x+\dfrac{5}{2})^2+\dfrac{3}{4}.$

1. Use the completed square form to justify the minimum possible value.

From part (a),
$x^2+5x+7=(x+\dfrac{5}{2})^2+\dfrac{3}{4}.$
Since $(x+\dfrac{5}{2})^2\geq 0$ for all real $x$,
$x^2+5x+7\geq \dfrac{3}{4}>0.$
Hence the expression will never be negative.