$$2x^3+5x^2+2x = 624$$
$$ \begin{matrix}x_1 = 6 & x_2 = -\dfrac{ 17 }{ 4 }+\dfrac{\sqrt{ 543 }}{ 4 }i & x_3 = -\dfrac{ 17 }{ 4 }-\dfrac{\sqrt{ 543 }}{ 4 }i \end{matrix} $$
$ \color{blue}{ 2x^{3}+5x^{2}+2x-624 } $ 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 ( 2 ) are 1 2 .The factors of the constant term (-624) are 1 2 3 4 6 8 12 13 16 24 26 39 48 52 78 104 156 208 312 624 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 13 }{ 1 } , ~ \pm \frac{ 13 }{ 2 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 26 }{ 1 } , ~ \pm \frac{ 26 }{ 2 } , ~ \pm \frac{ 39 }{ 1 } , ~ \pm \frac{ 39 }{ 2 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 52 }{ 1 } , ~ \pm \frac{ 52 }{ 2 } , ~ \pm \frac{ 78 }{ 1 } , ~ \pm \frac{ 78 }{ 2 } , ~ \pm \frac{ 104 }{ 1 } , ~ \pm \frac{ 104 }{ 2 } , ~ \pm \frac{ 156 }{ 1 } , ~ \pm \frac{ 156 }{ 2 } , ~ \pm \frac{ 208 }{ 1 } , ~ \pm \frac{ 208 }{ 2 } , ~ \pm \frac{ 312 }{ 1 } , ~ \pm \frac{ 312 }{ 2 } , ~ \pm \frac{ 624 }{ 1 } , ~ \pm \frac{ 624 }{ 2 } ~ $$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(6) = 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 - 6} $
$$ \frac{ 2x^{3}+5x^{2}+2x-624 }{ \color{blue}{ x - 6 } } = 2x^{2}+17x+104 $$Polynomial $ 2x^{2}+17x+104 $ can be used to find the remaining roots.
$ \color{blue}{ 2x^{2}+17x+104 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.