1. We need to solve $x|x|-5|x+2|+6=0$ and count the number of real roots.

2. Since absolute values are present, split according to the sign changes of $x$ and $x+2$.

The critical points are:

• $x=-2$
• $x=0$

So consider three intervals:

• $x\ge 0$
• $-2\le x<0$
• $x<-2$

Case 1: $x\ge 0$

Then $|x|=x,\qquad |x+2|=x+2$ So the equation becomes $x\cdot x-5(x+2)+6=0$ $x^2-5x-10+6=0$ $x^2-5x-4=0$

Now solve: $x=\frac{5\pm\sqrt{25+16}}{2}=\frac{5\pm\sqrt{41}}{2}$

Since this case requires $x\ge 0$, only $x=\frac{5+\sqrt{41}}{2}$ is valid.

So this case gives 1 root.

Case 2: $-2\le x<0$

Then $|x|=-x,\qquad |x+2|=x+2$ So $x|x|=x(-x)=-x^2$

Hence the equation becomes $-x^2-5(x+2)+6=0$ $-x^2-5x-10+6=0$ $-x^2-5x-4=0$ Multiply by $-1$: $x^2+5x+4=0$ $(x+1)(x+4)=0$ So the roots are $x=-1,-4$

Now check the interval $-2\le x<0$:

• $x=-1$ is valid
• $x=-4$ is not valid

So this case gives 1 root.

Case 3: $x<-2$

Then $|x|=-x,\qquad |x+2|=-(x+2)=-x-2$ So $x|x|=x(-x)=-x^2$

The equation becomes $-x^2-5(-x-2)+6=0$ $-x^2+5x+10+6=0$ $-x^2+5x+16=0$ Multiply by $-1$: $x^2-5x-16=0$

Solve: $x=\frac{5\pm\sqrt{25+64}}{2}=\frac{5\pm\sqrt{89}}{2}$

For this case we need $x<-2$.

• $\frac{5+\sqrt{89}}{2}>0$, invalid
• $\frac{5-\sqrt{89}}{2}$ is negative. Check if it is less than $-2$: since $\sqrt{89}>9$, $\frac{5-\sqrt{89}}{2}<\frac{5-9}{2}=-2$ so it is valid.

So this case gives 1 root.

4. Total number of real roots: $1+1+1=3$

Therefore, the number of real roots is $\boxed{3}$

So the correct option is B.