Solve for k in
$\displaystyle
4k-1=\sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$
$\displaystyle 4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
$\displaystyle 4(16k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
$\displaystyle 64k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
this is my wrong attempt.