Solve for $x$ in this equation How do I solve for $x$ algebraically?
$$\dfrac{x^2(x^2-1)}{x+3} = 12$$

Multiplying both sides by $x+3$ gives $x^2(x^2-1) = 12 (x+3)$. This can be re-written as
$x^4-x^2-12x-36 = 0$. A few guesses shows that $3$ and $-2$ are solutions, so you can factor these out to get a quadratic.
Explicitly, the equation becomes $(x-3)(x+2)(x^2+x+6) = 0$, hence the solutions are $3$, $-2$, and $\frac{1}{2}(-1 \pm i \sqrt{23}$).
(And, of course, we check that none of these are $-3$.)
