$$-x^3+25x^2-240x+900 = 0$$
$$ \begin{matrix}x_1 = 10 & x_2 = \dfrac{ 15 }{ 2 }+\dfrac{ 3 \sqrt{ 15}}{ 2 }i & x_3 = \dfrac{ 15 }{ 2 }-\dfrac{ 3 \sqrt{ 15}}{ 2 }i \end{matrix} $$
$ \color{blue}{ -x^{3}+25x^{2}-240x+900 } $ 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 factor of the leading coefficient ( -1 ) is 1 .The factors of the constant term (900) are 1 2 3 4 5 6 9 10 12 15 18 20 25 30 36 45 50 60 75 90 100 150 180 225 300 450 900 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 150 }{ 1 } , ~ \pm \frac{ 180 }{ 1 } , ~ \pm \frac{ 225 }{ 1 } , ~ \pm \frac{ 300 }{ 1 } , ~ \pm \frac{ 450 }{ 1 } , ~ \pm \frac{ 900 }{ 1 } ~ $$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(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}{x - 10} $
$$ \frac{ -x^{3}+25x^{2}-240x+900 }{ \color{blue}{ x - 10 } } = -x^{2}+15x-90 $$Polynomial $ -x^{2}+15x-90 $ can be used to find the remaining roots.
$ \color{blue}{ -x^{2}+15x-90 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.