For an equation of the form $x^2-6x=c$, rewrite the left side by adding $9$ to form $(x-3)^2$.

Both pairs of solutions are equally spaced around $3$. The greatest-to-least difference is twice whichever pair has the greater distance from $3$.

Algebra is faster because the repeated expression factors cleanly. After determining that the outer pair has distance $\sqrt{k+9}$ from $3$, use Desmos to verify the resulting equation.

Enter $y=2\sqrt{x+9}$ and $y=14$. The $x$-coordinate of their intersection gives the value of $k$.

Set $k$ to the intersection value and graph $y=(x^2-6x)^2-(k+6)(x^2-6x)+6k$. Its leftmost and rightmost $x$-intercepts should differ by $14$.

Let $u=x^2-6x$. The equation becomes $u^2-(k+6)u+6k=0$, which factors as $(u-6)(u-k)=0$. Therefore, $x^2-6x=6$ or $x^2-6x=k$.

Complete the square in each equation. From $x^2-6x=6$, we get $(x-3)^2=15$, so this pair is $3\pm\sqrt{15}$. From $x^2-6x=k$, we get $(x-3)^2=k+9$, so this pair is $3\pm\sqrt{k+9}$.

The two solution pairs are centered at $3$. Thus, the difference between the greatest and least solutions is twice the larger of the two distances from $3$. Since that difference is $14$, the larger distance is $7$. Because $\sqrt{15}<7$, the distance $\sqrt{k+9}$ must equal $7$.

Square $\sqrt{k+9}=7$ to get $k+9=49$. Therefore, $k=40$.