Let the mean and the variance of 20 observations ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots ,{\mathrm{x}}_{20}\text{ be }15\text{ and }9$, respectively. For $\mathrm{\alpha }\in \mathrm{R}$, if the mean of ${\left({\mathrm{x}}_1+\mathrm{\alpha }\right)}^2,{\left({\mathrm{x}}_2+\mathrm{\alpha }\right)}^2,\dots .,{\left({\mathrm{x}}_{20}+\mathrm{\alpha }\right)}^2$ is 178, then the square of the maximum value of $\alpha$ is equal to______.

Mean = I 5, variance = 9 
$\Rightarrow {\mathrm{\Sigma x}}_{\mathrm{i}}=15\times 20=300$        ....(i)
Also, $\frac{{\mathrm{\Sigma x}}_{\mathrm{i}}^2}{20}-{\left(\frac{{\mathrm{\Sigma x}}_{\mathrm{i}}}{20}\right)}^2=9$
$\Rightarrow {\mathrm{\Sigma x}}_{\mathrm{i}}^2=\left(9+(15{)}^2\right)\times 20=234\times 20=4680$
Also, $\frac{\mathrm{\Sigma }{\left({\mathrm{x}}_{\mathrm{i}}+\mathrm{\alpha }\right)}^2}{20}=178\Rightarrow \mathrm{\Sigma }{\left({\mathrm{x}}_{\mathrm{i}}+\mathrm{\alpha }\right)}^2=3560$
$\begin{array}{l}\Rightarrow {\mathrm{\Sigma x}}_{\mathrm{i}}^2+{\mathrm{\Sigma \alpha }}^2+2{\mathrm{\alpha \Sigma x}}_{\mathrm{i}}=3560\\ \Rightarrow 4680+20{\mathrm{\alpha }}^2+2\mathrm{\alpha }\times 300=3560\\ \Rightarrow 20{\mathrm{\alpha }}^2+600\mathrm{\alpha }+1120=0\Rightarrow {\mathrm{\alpha }}^2+30\mathrm{\alpha }+56=0\\ \Rightarrow (\mathrm{\alpha }+28)(\mathrm{\alpha }+2)=0\Rightarrow \mathrm{\alpha }=-28,-2\end{array}$
Thus, square of maximum value of $\mathrm{\alpha }$ is 4.