jmc

We begin by completing the square for $a^2+6a-7.$ We know that the binomial to be squared will be in terms of $a+b$ because the exponent for $a^2$ is 1.

By expanding $(a+b{)}^2$, we get $a^2+2ba+b^2$. We get that $2ba=6a$, so $b=3$, and it follows that $(a+3{)}^2=a^2+6a+9$.

Therefore, $a^2+6a-7=a^2+6a+9-16=(a+3{)}^2-16$. Since the square of a real number is at least zero, the minimum value of $a^2+6a-7$ is $0-16=\boxed{-16}$.