Find the value of $\displaystyle k$ for which the mean is equals to the standard deviation for the following quantities. $\displaystyle k-2$, $\displaystyle k$, $\displaystyle k$ and $\displaystyle 5k-2$ where $\displaystyle k$ is not negative.

Attempt

Mean= $\displaystyle 4k-1$

Standard deviation= $\displaystyle \sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$

Since mean=standard deviation, $\displaystyle 4k-1=\sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$

Facing problems with simplifying this... Need some help with simplifying it and solving for k.