$$300x^4+645x^3+\frac{183}{2}x^2+3x = 0$$
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 10 } & x_3 = -\dfrac{ 1 }{ 20 } \\[1 em] x_4 = -2 & \\[1 em] \end{matrix} $$
In order to solve $ \color{blue}{ 600x^{4}+1290x^{3}+183x^{2}+6x = 0 } $, first we need to factor our $ x $.
$$ 600x^{4}+1290x^{3}+183x^{2}+6x = x \left( 600x^{3}+1290x^{2}+183x+6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 600x^{3}+1290x^{2}+183x+6 = 0$.
$ \color{blue}{ 600x^{3}+1290x^{2}+183x+6 } $ 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 ( 600 ) are 1 2 3 4 5 6 8 10 12 15 20 24 25 30 40 50 60 75 100 120 150 200 300 600 .The factors of the constant term (6) are 1 2 3 6 . 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 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 12 } , ~ \pm \frac{ 1 }{ 15 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 1 }{ 24 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 30 } , ~ \pm \frac{ 1 }{ 40 } , ~ \pm \frac{ 1 }{ 50 } , ~ \pm \frac{ 1 }{ 60 } , ~ \pm \frac{ 1 }{ 75 } , ~ \pm \frac{ 1 }{ 100 } , ~ \pm \frac{ 1 }{ 120 } , ~ \pm \frac{ 1 }{ 150 } , ~ \pm \frac{ 1 }{ 200 } , ~ \pm \frac{ 1 }{ 300 } , ~ \pm \frac{ 1 }{ 600 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 2 }{ 8 } , ~ \pm \frac{ 2 }{ 10 } , ~ \pm \frac{ 2 }{ 12 } , ~ \pm \frac{ 2 }{ 15 } , ~ \pm \frac{ 2 }{ 20 } , ~ \pm \frac{ 2 }{ 24 } , ~ \pm \frac{ 2 }{ 25 } , ~ \pm \frac{ 2 }{ 30 } , ~ \pm \frac{ 2 }{ 40 } , ~ \pm \frac{ 2 }{ 50 } , ~ \pm \frac{ 2 }{ 60 } , ~ \pm \frac{ 2 }{ 75 } , ~ \pm \frac{ 2 }{ 100 } , ~ \pm \frac{ 2 }{ 120 } , ~ \pm \frac{ 2 }{ 150 } , ~ \pm \frac{ 2 }{ 200 } , ~ \pm \frac{ 2 }{ 300 } , ~ \pm \frac{ 2 }{ 600 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 3 }{ 10 } , ~ \pm \frac{ 3 }{ 12 } , ~ \pm \frac{ 3 }{ 15 } , ~ \pm \frac{ 3 }{ 20 } , ~ \pm \frac{ 3 }{ 24 } , ~ \pm \frac{ 3 }{ 25 } , ~ \pm \frac{ 3 }{ 30 } , ~ \pm \frac{ 3 }{ 40 } , ~ \pm \frac{ 3 }{ 50 } , ~ \pm \frac{ 3 }{ 60 } , ~ \pm \frac{ 3 }{ 75 } , ~ \pm \frac{ 3 }{ 100 } , ~ \pm \frac{ 3 }{ 120 } , ~ \pm \frac{ 3 }{ 150 } , ~ \pm \frac{ 3 }{ 200 } , ~ \pm \frac{ 3 }{ 300 } , ~ \pm \frac{ 3 }{ 600 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 6 }{ 5 } , ~ \pm \frac{ 6 }{ 6 } , ~ \pm \frac{ 6 }{ 8 } , ~ \pm \frac{ 6 }{ 10 } , ~ \pm \frac{ 6 }{ 12 } , ~ \pm \frac{ 6 }{ 15 } , ~ \pm \frac{ 6 }{ 20 } , ~ \pm \frac{ 6 }{ 24 } , ~ \pm \frac{ 6 }{ 25 } , ~ \pm \frac{ 6 }{ 30 } , ~ \pm \frac{ 6 }{ 40 } , ~ \pm \frac{ 6 }{ 50 } , ~ \pm \frac{ 6 }{ 60 } , ~ \pm \frac{ 6 }{ 75 } , ~ \pm \frac{ 6 }{ 100 } , ~ \pm \frac{ 6 }{ 120 } , ~ \pm \frac{ 6 }{ 150 } , ~ \pm \frac{ 6 }{ 200 } , ~ \pm \frac{ 6 }{ 300 } , ~ \pm \frac{ 6 }{ 600 } ~ $$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(-\frac{ 1 }{ 10 }) = 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}{ 10 x + 1 } $
$$ \frac{ 600x^{3}+1290x^{2}+183x+6 }{ \color{blue}{ 10x + 1 } } = 60x^{2}+123x+6 $$Polynomial $ 60x^{2}+123x+6 $ can be used to find the remaining roots.
$ \color{blue}{ 60x^{2}+123x+6 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.