$$(3x^2-24)(x^2+2x) = 0$$
$$ \begin{matrix}x_1 = 0 & x_2 = -2 & x_3 = -2 \sqrt{ 2 } \\[1 em] x_4 = 2 \sqrt{ 2 } & \\[1 em] \end{matrix} $$
In order to solve $ \color{blue}{ 3x^{4}+6x^{3}-24x^{2}-48x = 0 } $, first we need to factor our $ x $.
$$ 3x^{4}+6x^{3}-24x^{2}-48x = x \left( 3x^{3}+6x^{2}-24x-48 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{3}+6x^{2}-24x-48 = 0$.
$ \color{blue}{ 3x^{3}+6x^{2}-24x-48 } $ is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.
The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.
The factors of the leading coefficient ( 3 ) are 1 3 .The factors of the constant term (-48) are 1 2 3 4 6 8 12 16 24 48 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 3 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 3 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 3 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 3 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 3 } ~ $$Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
If we plug these values into the polynomial $ P(x) $, we obtain $ P(-2) = 0 $.
To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:
Divide $ P(x) $ with $ \color{blue}{x + 2} $
$$ \frac{ 3x^{3}+6x^{2}-24x-48 }{ \color{blue}{ x + 2 } } = 3x^{2}-24 $$Polynomial $ 3x^{2}-24 $ can be used to find the remaining roots.
$ \color{blue}{ 3x^{2}-24 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.