Answer
The trinomial $x^{2}+7x+c$ is a perfect square when $c=\displaystyle \frac{49}{4}.$
Then $x^{2}+7x+\displaystyle \frac{49}{4}=(x+\frac{7}{2})(x+\frac{7}{2})=(x+\frac{7}{2})^{2}$
Work Step by Step
$ x^{2}+7x+c\qquad$ ... find half the coefficient of $x$, which is $\displaystyle \frac{7}{2}$.
$\qquad$ ...square the result.
$(\displaystyle \frac{7}{2})^{2}=\frac{49}{4}\qquad$ ...substitute $c$ with $\displaystyle \frac{49}{4}$ in the original expression
$x^{2}+7x+\displaystyle \frac{49}{4}=(x+\frac{7}{2})^{2}$
