Question 387384
I can say {{{k = b^2}}}
The equation becomes
{{{k^2 + 13k + 36 = 0}}}
Using the quadratic formula:
{{{k = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
where
{{{a = 1}}}
{{{b = 13}}}
{{{c = 36}}}
{{{k = (-13 +- sqrt( 13^2-4*1*36 ))/(2*1) }}} 
{{{k = (-13 +- sqrt( 169 - 144 ))/2 }}}
{{{k = (-13 +- sqrt( 25 ))/2 }}}
{{{k = (-13 + 5)/2}}}
{{{k = -4}}}
and
{{{k = (-13 - 5)/2}}}
{{{k = -9}}}
And, since
{{{k = b^2}}}
{{{-4 = b^2}}}
{{{b = 2i}}}
and
{{{-9 = b^2}}}
{{{b = 3i}}}
The roots are imaginary
If I put {{{b = 3i}}} back into the equation:
{{{(3i)^4 + 13*(3i)^2 + 36 = 0}}}
{{{81 - 117 + 36 = 0}}}
{{{117 - 117 = 0}}}
{{{b = 2i}}} should check also
(note that{{{ i^4 = 1}}})