Answer
$$-17$$
Work Step by Step
Recall that the definition $\sqrt{-a} = i\sqrt{a}$ must be applied BEFORE the other rules regarding radicals, therefore,
$\sqrt{-17}*\sqrt{-17} =i\sqrt{17}*i\sqrt{17} $
now we may multiply
$i^2(17)$
$(-1)(17)$
$$-17$$
which checks out because $(\sqrt{a})^2=a$
