We need to find the number of real solutions of $x|x+5|+2|x+7|-2=0.$

We solve it by breaking into cases according to the absolute values.

1. Critical points

The expressions inside modulus are:

• $x+5=0 \Rightarrow x=-5$
• $x+7=0 \Rightarrow x=-7$

So we consider three intervals:

1. $x\ge -5$
2. $-7\le x< -5$
3. $x< -7$

2. Case I: $x\ge -5$

Then $|x+5|=x+5,\qquad |x+7|=x+7.$ So the equation becomes $x(x+5)+2(x+7)-2=0.$ Simplifying, $x^2+5x+2x+14-2=0$ $x^2+7x+12=0.$ Factorizing, $(x+3)(x+4)=0.$ Thus, $x=-3,-4.$ Both satisfy $x\ge -5$, so both are valid.

So from this case, we get 2 solutions.

3. Case II: $-7\le x< -5$

Then $|x+5|=-(x+5),\qquad |x+7|=x+7.$ So the equation becomes $x[-(x+5)] +2(x+7)-2=0.$ That is, $-x^2-5x+2x+14-2=0$ $-x^2-3x+12=0.$ Multiply by $-1$: $x^2+3x-12=0.$ Using the quadratic formula, $x=\frac{-3\pm\sqrt{9+48}}{2}=\frac{-3\pm\sqrt{57}}{2}.$ Now check which root lies in $[-7,-5)$:

• $\frac{-3+\sqrt{57}}{2}>0$, not valid.
• $\frac{-3-\sqrt{57}}{2}\approx \frac{-3-7.55}{2}\approx -5.27,$ which lies in $[-7,-5)$, so it is valid.

So from this case, we get 1 solution.

4. Case III: $x< -7$

Then $|x+5|=-(x+5),\qquad |x+7|=-(x+7).$ So the equation becomes $x[-(x+5)] +2[-(x+7)]-2=0.$ Thus, $-x^2-5x-2x-14-2=0$ $-x^2-7x-16=0.$ Multiply by $-1$: $x^2+7x+16=0.$ Discriminant: $\Delta=7^2-4\cdot 1\cdot 16=49-64=-15<0.$ So there is no real solution in this case.

5. Total number of real solutions

Adding all valid solutions: $2+1+0=3.$

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

6. Comparison with stored answer

Stored correct answer: $3$

Our derived answer is also $3$, so it agrees.