If we let R be the value to the left of M and S be the value to the right of M, then we can write the equation $M = \sqrt{S}-\sqrt{R}$. Substituting in $M=\sqrt{270}$, we have $\sqrt{270} = \sqrt{S}-\sqrt{R}$. Squaring both sides, we get $270 = S+\cancel{R}-2\sqrt{RS}$, or $130 = S-R-2\sqrt{RS}$. Rearranging further, we have $R+S-130 = -2(\sqrt{RS})$. Squaring both sides once more, we get $(R+S-130)^2 = 4RS$. Letting $A=R+S$, we have $A=130$, and our equation becomes $A^2 - 260A + 16900 = 4RS$. We are given that the values of R and S are decimal, so both must be positive since they exist on the number line. We also know that R and S must satisfy the equation $RS = \left(\dfrac{A^2 -260A+16900}{4}\right)$, or $RS = \left(\dfrac{A^2}{4} -65A+\dfrac{16900}{4}\right)$. To maximize RS, we must maximize A, so we have $A = 130$. Plugging this into our equation for RS gives $RS = \left(4225-8450+\dfrac{16900}{4}\right)$, or $RS = \left(\dfrac{4225}{4}\right)$. To find the decimal values of R and S, we must divide $\dfrac{4225}{4} \approx \boxed{1056.3}$.
