For what interval of $k$ does the equation have one postitive and one negative root? I was given an equation that looked like this 
$(k-x)(1-x) + 4$
and was told to find the interval of k when the equation would have one positive root and one negative root.
So far I have found the solutions in terms of k 
$x_1 = \frac{1}{2}(k+1-\sqrt{k^2-2k-15})$
$x_2 = \frac{1}{2}(k+1+\sqrt{k^2-2k-15})$
I am a bit stuck of where to go from here, do we have to figure it out using 
$\sqrt{b^2-4ac}$?

I made some silly errors
in my original answer.
The corrections now seem to show that
k < -4 is the condition.
Following dxiv's advice,
the roots are
$x_1 = \frac{1}{2}(-\sqrt(k^2-2k-15)+k+1)
$
and
$x_2 = \frac{1}{2}(\sqrt(k^2-2k-15)+k+1)
$.
For there to be
two real roots,
we must have
$k^2-2k-15 > 0$
or
$k^2-2k+1 > 16$
or
$(k-1)^2 > 16
$
or
$k-1 > 4$
or
$k-1 < -4$
or
$k > 5$
or
$k < -3$.
(Error - had -5 here)
To make the roots
of different signs,
consider the cases separately.
If $k > 5$
then
$\sqrt(k^2-2k-15)+k+1 > 0$
so we want
$-\sqrt(k^2-2k-15)+k+1
\lt 0$
or
$\sqrt(k^2-2k-15)
\gt k+1
$
or
$k^2-2k-15
\gt k^2+2k+1
$
or
$0 > 4k+16$
which never happens.
(I messed up the inequality below
also)
If $k < -3$
then
$\sqrt(k^2-2k-15)+k+1 > 0$
so we want
$\sqrt(k^2-2k-15)
\gt -k-1
$
or
$k^2-2k-15
\gt k^2+2k+1
$
or
$0 > 4k+16$
or
$k <-4$,
