Question 1019225
<pre>
{{{x^2-(k+2)x+k+5}}}{{{""=""}}}{{{0}}}

The solution using the quadratic formula:

{{{(-(-(k+2)) +- sqrt( (-(k+2))^2-4*(1)*(k+5) ))/(2*(1)) <0 }}}

simplifies to

{{{(k+2 +- sqrt(k^2-16))/2}}}

We only need to require that the larger
solution, the one with the plus sign before
the square root, be negative, so

{{{(k+2 + sqrt(k^2-16))/2<0}}} 

The discriminant must be positive to prevent
getting 1 single or 2 imaginary solutions, so 
setting the discriminant greater than 0,

{{{k^2-16>0}}}

That has solutions k < -4 or k > 4.
If k > 4 then certainly the larger solution 
will be positive.

So we must take k < -4

{{{(k+2 + sqrt(k^2-16))/2<0}}}

Multiplying both sides by 2

{{{k+2 + sqrt(k^2-16)<0}}}

{{{sqrt(k^2-16)<-k-2}}}

{{{k^2-16 <k^2+4k+4}}}

{{{-16 <4k+4}}}

{{{-20 <4k}}}

{{{-5<k}}}

Therefore to have both solutions negative,

{{{-5<k<-4}}}

Edwin</pre>