Find how many solutions has the following equation... 
Determine how many real solutions has the following equation:

$$x^2(|x|-6)=-15$$
I noticed that $|x|-6$ should be negative because $x^2$ is always a positive value. Thus, $x\in(-6;6)$. I made a substitution: $|x|=t$, hence $x^2=t^2$, and got the equation:$$t^3-6t^2+15=0$$
Now the problem is I cannot find any solution for this equation. Hope you'll give me the right   explanation for this exercise. Thank you very much!
 A: One way to answer this question is to draw a graph of $y = x^2(|x| - 6)$, then draw a horizontal line at $y = -15$.  The curve and the line obviously intersect in four places. QED.
Finding the solutions is a bit trickier, but that is not the question that was asked.
If you don't want to rely on drawing the graph, you can prove the result using $f(t)$ and $f'(t)$.  As you showed, $f(t) = t^3 - 6t^2 + 15$, which gives $f'(t) = 3t^2 - 12t$.
Solving for $f'(t) = 0$ yields $t = \{0, 4\}.$  Thus, $f(t)$ has extrema at $t =0$ and $t = 4$.  We only care about $t>0$, so we ignore that one.  Calculating $f'(1) = -9$ and $f'(5) = 15$, we see that $f(t)$ must be strictly decreasing for $0 < t < 4$ and strictly increasing for $t > 4$.
Some key evaluations:
$$
f(0) = 15
$$
$$
f(4) = -17
$$
$$
f(6) = 15
$$
Since $f(t)$ is strictly decreasing between $0$ and $4$, and $f(0) > 0$ and $f(4) < 0$, $f(t)$ must cross zero exactly once in that region.  Likewise, since $f(t)$ is strictly increasing for $t > 4$ and $f(4) < 0$ and $f(6) > 0$, $f(t)$ must cross zero exactly once in that region.
Thus, there are two positive solutions to the initial equation.  By symmetry, there must be two negative solutions, as well.
