$$x^4+10x^3+50x^2+125x-6006 = 0$$
$$ \begin{matrix}x_1 = 6 & x_2 = -11 & x_3 = -\dfrac{ 5 }{ 2 }+\dfrac{\sqrt{ 339 }}{ 2 }i \\[1 em] x_4 = -\dfrac{ 5 }{ 2 }-\dfrac{\sqrt{ 339 }}{ 2 }i & \\[1 em] \end{matrix} $$
$ \color{blue}{ x^{4}+10x^{3}+50x^{2}+125x-6006 } $ is a polynomial of degree 4. 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (-6006) are 1 2 3 6 7 11 13 14 21 22 26 33 39 42 66 77 78 91 143 154 182 231 273 286 429 462 546 858 1001 2002 3003 6006 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 11 }{ 1 } , ~ \pm \frac{ 13 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 22 }{ 1 } , ~ \pm \frac{ 26 }{ 1 } , ~ \pm \frac{ 33 }{ 1 } , ~ \pm \frac{ 39 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 66 }{ 1 } , ~ \pm \frac{ 77 }{ 1 } , ~ \pm \frac{ 78 }{ 1 } , ~ \pm \frac{ 91 }{ 1 } , ~ \pm \frac{ 143 }{ 1 } , ~ \pm \frac{ 154 }{ 1 } , ~ \pm \frac{ 182 }{ 1 } , ~ \pm \frac{ 231 }{ 1 } , ~ \pm \frac{ 273 }{ 1 } , ~ \pm \frac{ 286 }{ 1 } , ~ \pm \frac{ 429 }{ 1 } , ~ \pm \frac{ 462 }{ 1 } , ~ \pm \frac{ 546 }{ 1 } , ~ \pm \frac{ 858 }{ 1 } , ~ \pm \frac{ 1001 }{ 1 } , ~ \pm \frac{ 2002 }{ 1 } , ~ \pm \frac{ 3003 }{ 1 } , ~ \pm \frac{ 6006 }{ 1 } ~ $$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} $
$$ \frac{ x^{4}+10x^{3}+50x^{2}+125x-6006 }{ \color{blue}{ x - 6 } } = x^{3}+16x^{2}+146x+1001 $$Polynomial $ x^{3}+16x^{2}+146x+1001 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+16x^{2}+146x+1001 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.