Find the values of $k$ for the following quadratic equation,so that it has two equal roots:
$k x(x-2)+6=0$

(K=6) Given equation: $k x(x-2)+6=0$
Expanding the equation: $k x^{2}-2 k x+6=0$
Comparing this with the standard quadratic form $a x^{2}+b x+c=0$,we get:
$a=k, b=-2 k, c=6$
For a quadratic equation to have two equal roots,the discriminant $D$ must be equal to $0$:
$D = b^{2}-4 a c = 0$
Substituting the values:
$(-2 k)^{2}-4(k)(6) = 0$
$4 k^{2}-24 k = 0$
Factoring out $4k$:
$4 k(k-6) = 0$
This gives two possibilities: $4 k=0$ or $k-6=0$,which implies $k=0$ or $k=6$.
However,if $k=0$,the equation becomes $6=0$,which is not a quadratic equation. Thus,$k$ cannot be $0$.
Therefore,the only valid value is $k=6$.