If $\displaystyle \bold{a}=<k,2>$ and $\displaystyle \bold{b}=<7,6>$, then $\displaystyle \bold{a}-\bold{b}=<k-7,2-6>=<k-7,-4>$. Therefore:

$\displaystyle {\left|\bold{a}-\bold{b}\right|}=\sqrt{(k-7)^2+(-4)^2}=\sqrt{k^2-14k+49+16}=\sqrt{k^2-14k+65}$

Now find where

$\displaystyle {\left|\bold{a}-\bold{b}\right|}=5$.

$\displaystyle 5=\sqrt{k^2-14k+65}$

$\displaystyle \rightarrow 25=k^2-14k+65$

$\displaystyle \rightarrow k^2-14k+40=0$

$\displaystyle (k-4)(k-10)=0$

$\displaystyle \rightarrow k=4$ or $\displaystyle k=10$.

You can plug these back in to verify the k values.

$\displaystyle 5=\sqrt{4^2-14(4)+65}=\sqrt{16-56+65}=\sqrt{25}={\color{red}5}$

$\displaystyle 5=\sqrt{10^2-14(10)+65}=\sqrt{100-140+65}=\sqrt{25}={\color{red}5}$

Hope that answers your question!!