1. Let $x=a-b$ where $(a,b)\in A\times A$ and $A=\{1,2,3,\dots,10\}$.

Then the condition for the relation becomes $2(a-b)^2+3(a-b)\in B=\{0,1,2,3,4\}$ which is $2x^2+3x\in\{0,1,2,3,4\}.$

2. Since $a,b\in\{1,2,\dots,10\}$, the difference $x=a-b$ can take integer values from $-9$ to $9$.

We now check for which integers $x$, $2x^2+3x\in\{0,1,2,3,4\}.$

3. Solve separately for each possible value in $B$.

We test small integer values of $x$:

• For $x=0$: $2(0)^2+3(0)=0\in B$
• For $x=1$: $2(1)^2+3(1)=2+3=5\notin B$
• For $x=-1$: $2(1)+3(-1)=2-3=-1\notin B$
• For $x=2$: $2(4)+6=14\notin B$
• For $x=-2$: $2(4)-6=2\in B$

For $|x|\ge 3$, the value $2x^2+3x$ is too large positive or less than $0$ except no new values in $\{0,1,2,3,4\}$ occur. So the only possible differences are $x=0,-2.$

4. Count ordered pairs $(a,b)$ for each allowed difference.

Case 1: $a-b=0$

Then $a=b$. Since $a,b\in A=\{1,2,\dots,10\}$, there are $10$ pairs: $(1,1),(2,2),\dots,(10,10).$

Case 2: $a-b=-2$

Then $a=b-2\quad\text{or}\quad b=a+2.$ To keep both $a,b\in\{1,2,\dots,10\}$, we can choose $a=1,2,\dots,8$ which gives pairs $(1,3),(2,4),\dots,(8,10).$ So the number of such pairs is $8.$

5. Total number of elements in $R$ is $10+8=18.$

Therefore, the required number of elements is $\boxed{18}.$

6. Comparison with stored answer:

Stored correct answer = $18$.

This matches our derived answer.