$$0 = 18x^4+87x^3+3x^2-108x$$
$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = -\dfrac{ 4 }{ 3 } \\[1 em] x_4 = -\dfrac{ 9 }{ 2 } & \\[1 em] \end{matrix} $$
In order to solve $ \color{blue}{ -18x^{4}-87x^{3}-3x^{2}+108x = 0 } $, first we need to factor our $ x $.
$$ -18x^{4}-87x^{3}-3x^{2}+108x = x \left( -18x^{3}-87x^{2}-3x+108 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -18x^{3}-87x^{2}-3x+108 = 0$.
$ \color{blue}{ -18x^{3}-87x^{2}-3x+108 } $ 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 ( -18 ) are 1 2 3 6 9 18 .The factors of the constant term (108) are 1 2 3 4 6 9 12 18 27 36 54 108 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 18 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 2 }{ 9 } , ~ \pm \frac{ 2 }{ 18 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 3 }{ 9 } , ~ \pm \frac{ 3 }{ 18 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 6 } , ~ \pm \frac{ 4 }{ 9 } , ~ \pm \frac{ 4 }{ 18 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 6 }{ 6 } , ~ \pm \frac{ 6 }{ 9 } , ~ \pm \frac{ 6 }{ 18 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 3 } , ~ \pm \frac{ 9 }{ 6 } , ~ \pm \frac{ 9 }{ 9 } , ~ \pm \frac{ 9 }{ 18 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 3 } , ~ \pm \frac{ 12 }{ 6 } , ~ \pm \frac{ 12 }{ 9 } , ~ \pm \frac{ 12 }{ 18 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 18 }{ 3 } , ~ \pm \frac{ 18 }{ 6 } , ~ \pm \frac{ 18 }{ 9 } , ~ \pm \frac{ 18 }{ 18 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 27 }{ 3 } , ~ \pm \frac{ 27 }{ 6 } , ~ \pm \frac{ 27 }{ 9 } , ~ \pm \frac{ 27 }{ 18 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 36 }{ 3 } , ~ \pm \frac{ 36 }{ 6 } , ~ \pm \frac{ 36 }{ 9 } , ~ \pm \frac{ 36 }{ 18 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } , ~ \pm \frac{ 54 }{ 3 } , ~ \pm \frac{ 54 }{ 6 } , ~ \pm \frac{ 54 }{ 9 } , ~ \pm \frac{ 54 }{ 18 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 108 }{ 2 } , ~ \pm \frac{ 108 }{ 3 } , ~ \pm \frac{ 108 }{ 6 } , ~ \pm \frac{ 108 }{ 9 } , ~ \pm \frac{ 108 }{ 18 } ~ $$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(1) = 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 - 1} $
$$ \frac{ -18x^{3}-87x^{2}-3x+108 }{ \color{blue}{ x - 1 } } = -18x^{2}-105x-108 $$Polynomial $ -18x^{2}-105x-108 $ can be used to find the remaining roots.
$ \color{blue}{ -18x^{2}-105x-108 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.