Question 664293
You are told to find k such that the given equation results in two equal roots.
That says the resulting answer will end up being in the form {{{(ax+b)^2 = 0}}}
So Let's expand that 
 {{{(ax+b)^2 = 0}}}
{{{a^2x^2 + 2abx + b^2 = 0}}}
Now lets match that up with the given problem
We see that {{{k = a^2}}}, {{{12 = 2ab}}}, and {{{k = b^2}}}
We can see from this that {{{6=ab}}} and also that {{{a^2 = b^2}}}
Using the {{{a^2 = b^2}}}, we can see that either a=b or a=-b
Since a*b is positive (6), then a must be equal to b.

If a=b and ab =6, then a and b are both {{{sqrt(6)}}}
So what does than make k? k is a^2 = 6.

Check our answer
{{{kx^2+12x+k=0}}}
{{{6x^2 + 12x + 6 = 0}}}
{{{x^2 +2x +1 = 0}}}
{{{(x+1)^2 = 0}}}