Since you asked to solve that equation, I would assume you need a method that can be used to get the answer to any arbitrary precision. Since Newton's method is usually not taught in high school, let us try an alternative that does not require anything like gradients or even square roots.
$x^3+3 x^2+1=0\Leftrightarrow x^2(x+3)=-1\Leftrightarrow x=-1/x^2-3$
Now the problem becomes finding $x$ so that it equals $-1/x^2-3$.
One might be very quick to find out that $x<3$ and then plug that inequality back into the right hand side and get $x>-3-1/9$. Then you can plug that back into RHS again and get $x<-2433/784\approx-3.10332$, then do it again you get $x>-(18373123/5919489)\approx-3.10383599$. You can do it over and over again to get tighter and tighter upper and lower bounds. You can get a rational bound to arbitrary precision here.
But what if one is from one of those high school that doesn't even teach inequalities?
Well the worst one can do is to guess an arbitrary number (other than 0, of course) for $x$, you can evaluate this and see if actually $x$ do equal to $-1/x^2-3$.
Chances are, you won't be that lucky and they just happen to be the same; maybe they are even orders of magnitudes away. Well, two things that are supposed to be equal are not so, then they probably should be between that. In other words, obtain a better guess by taking the average of these two numbers:
Now we have a better new guess, and keep the procedure you can refine that result to arbitrary precision.
In fact, even better than Newton's method, this kind of iterative procedure converges from any starting guess (other than 0), and the concept is extremely easy. Of course, you have to carry out the work of iterations.