Answer
$-8\lt x\lt2$
Work Step by Step
$x^2-6x-16=(x-2)(x+8)$. The coefficient of $x^2$ is 1, hence the coefficients of $x$ in the multiplicants are $1$.
$(x-2)(x+8)\lt0$, the leading coefficient of the function is $1$, therefore it is open up, thus it is negative when $-8\lt x\lt2$. (See the graph of the function, the solution is the area under the x-axis.)
