Step 1: Use the repeated root idea.
The quartic has two distinct roots, each repeated, so it equals $(x-a)^2(x-b)^2$. This is the same as $\big(x^2-(a+b)x+ab\big)^2$.
Step 2: Compare with the given polynomial.
Expanding the square and matching with $x^4-10x^3+37x^2-60x+36$, the $x^3$ term gives $2(a+b)=10$, so $a+b=5$.
Step 3: Match the constant term.
The constant is $(ab)^2=36$, so $ab=6$ (taking the positive value that fits).
Step 4: Find the sum of squares of $a$ and $b$.
Use $a^2+b^2=(a+b)^2-2ab=25-12=13$.
Step 5: Account for repetition.
The four roots are $a,a,b,b$, so the sum of squares of all roots is $2a^2+2b^2=2(a^2+b^2)$.
Step 6: Compute the final value.
That is $2\times13=26$. \[ \boxed{26} \]