1. Given function

We need the points in $[-1,2]$ where $f(x)=2x^2+x+[x^2]-[x]$ is not continuous.

Here, $2x^2+x$ is a polynomial, so it is continuous everywhere. Hence discontinuity can only come from $g(x)=[x^2]-[x].$ So we only need to check where $[x^2]$ or $[x]$ changes value.

2. Possible discontinuity points

(i) For $[x]$

The greatest integer function $[x]$ is discontinuous at every integer. In $[-1,2]$, the relevant integers are $x=-1,0,1,2.$ For continuity inside the interval, we mainly check interior points $0,1$ and also endpoint behavior if needed. Since the question asks number of points where $f$ is not continuous on the interval, endpoints can also be checked via one-sided continuity.

(ii) For $[x^2]$

This changes when $x^2$ crosses an integer. In $[-1,2]$, we have $x^2\in[0,4]$, so possible jumps occur when $x^2=1,2,3,4.$ Thus possible points are $x=\pm 1,\ \pm\sqrt2,\ \pm\sqrt3,\ \pm 2.$ Restricting to $[-1,2]$, we get $x=-1,1,\sqrt2,\sqrt3,2.$

Combining both lists, the only possible discontinuity points are $x=-1,0,1,\sqrt2,\sqrt3,2.$

3. Check each candidate point

We analyze $g(x)=[x^2]-[x]$.

(a) At $x=0$

For $x\to 0^-$:

• $[x]=-1$
• $x^2\in(0,1)$, so $[x^2]=0$ Thus $g(x)=0-(-1)=1.$

For $x\to 0^+$:

• $[x]=0$
• $[x^2]=0$ Thus $g(x)=0-0=0.$

Left and right limits differ, so $f$ is discontinuous at $x=0$.

(b) At $x=1$

For $x\to 1^-$:

• $[x]=0$
• $x^2\to 1^-$, so $[x^2]=0$ Thus $g(x)=0-0=0.$

At $x=1$ and for $x\to 1^+$:

• $[x]=1$
• for $x>1$ close to $1$, $x^2\in(1,2)$ so $[x^2]=1$ Thus $g(x)=1-1=0.$ Also at $x=1$, $g(1)=1-1=0$.

Hence $f$ is continuous at $x=1$.

(c) At $x=\sqrt2$

Here $1<\sqrt2<2$.

For $x\to \sqrt2^-$:

• $[x]=1$
• $x^2<2$, so $[x^2]=1$ Thus $g(x)=1-1=0.$

For $x\to \sqrt2^+$:

• $[x]=1$
• $x^2>2$, so $[x^2]=2$ Thus $g(x)=2-1=1.$

Jump occurs, so discontinuous at $x=\sqrt2$.

(d) At $x=\sqrt3$

Again $1<\sqrt3<2$.

For $x\to \sqrt3^-$:

• $[x]=1$
• $x^2<3$, so $[x^2]=2$ Thus $g(x)=2-1=1.$

For $x\to \sqrt3^+$:

• $[x]=1$
• $x^2>3$, so $[x^2]=3$ Thus $g(x)=3-1=2.$

Jump occurs, so discontinuous at $x=\sqrt3$.

(e) At $x=-1$

Check right continuity since it is the left endpoint. For $x>-1$ close to $-1$:

• $[x]=-1$
• $x^2<1$, so $[x^2]=0$ Thus $g(x)=0-(-1)=1.$

But at $x=-1$: $g(-1)=[1]-[-1]=1-(-1)=2.$ So right limit is not equal to function value. Hence $f$ is discontinuous at $x=-1$.

(f) At $x=2$

Check left continuity since it is the right endpoint. For $x<2$ close to $2$:

• $[x]=1$
• $x^2<4$ but close to $4$, so $[x^2]=3$ Thus $g(x)=3-1=2.$

At $x=2$: $g(2)=[4]-[2]=4-2=2.$ So left limit equals function value. Hence $f$ is continuous at $x=2$.

4. List of discontinuity points

The discontinuities occur at $x=-1,\ 0,\ \sqrt2,\ \sqrt3.$ So the number of points is $4.$

5. Option matching

Thus the correct option is $\boxed{\text{C: }4}.$