$$-15625x^3+31250x^2-17175x+2678 = 0$$
$$ \begin{matrix}x_1 = \dfrac{ 13 }{ 25 } & x_2 = \dfrac{ 37 }{ 50 }-\dfrac{\sqrt{ 545 }}{ 50 } & x_3 = \dfrac{ 37 }{ 50 }+\dfrac{\sqrt{ 545 }}{ 50 } \end{matrix} $$
$ \color{blue}{ -15625x^{3}+31250x^{2}-17175x+2678 } $ 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 ( -15625 ) are 1 5 25 125 625 3125 15625 .The factors of the constant term (2678) are 1 2 13 26 103 206 1339 2678 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 1 }{ 625 } , ~ \pm \frac{ 1 }{ 3125 } , ~ \pm \frac{ 1 }{ 15625 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 2 }{ 25 } , ~ \pm \frac{ 2 }{ 125 } , ~ \pm \frac{ 2 }{ 625 } , ~ \pm \frac{ 2 }{ 3125 } , ~ \pm \frac{ 2 }{ 15625 } , ~ \pm \frac{ 13 }{ 1 } , ~ \pm \frac{ 13 }{ 5 } , ~ \pm \frac{ 13 }{ 25 } , ~ \pm \frac{ 13 }{ 125 } , ~ \pm \frac{ 13 }{ 625 } , ~ \pm \frac{ 13 }{ 3125 } , ~ \pm \frac{ 13 }{ 15625 } , ~ \pm \frac{ 26 }{ 1 } , ~ \pm \frac{ 26 }{ 5 } , ~ \pm \frac{ 26 }{ 25 } , ~ \pm \frac{ 26 }{ 125 } , ~ \pm \frac{ 26 }{ 625 } , ~ \pm \frac{ 26 }{ 3125 } , ~ \pm \frac{ 26 }{ 15625 } , ~ \pm \frac{ 103 }{ 1 } , ~ \pm \frac{ 103 }{ 5 } , ~ \pm \frac{ 103 }{ 25 } , ~ \pm \frac{ 103 }{ 125 } , ~ \pm \frac{ 103 }{ 625 } , ~ \pm \frac{ 103 }{ 3125 } , ~ \pm \frac{ 103 }{ 15625 } , ~ \pm \frac{ 206 }{ 1 } , ~ \pm \frac{ 206 }{ 5 } , ~ \pm \frac{ 206 }{ 25 } , ~ \pm \frac{ 206 }{ 125 } , ~ \pm \frac{ 206 }{ 625 } , ~ \pm \frac{ 206 }{ 3125 } , ~ \pm \frac{ 206 }{ 15625 } , ~ \pm \frac{ 1339 }{ 1 } , ~ \pm \frac{ 1339 }{ 5 } , ~ \pm \frac{ 1339 }{ 25 } , ~ \pm \frac{ 1339 }{ 125 } , ~ \pm \frac{ 1339 }{ 625 } , ~ \pm \frac{ 1339 }{ 3125 } , ~ \pm \frac{ 1339 }{ 15625 } , ~ \pm \frac{ 2678 }{ 1 } , ~ \pm \frac{ 2678 }{ 5 } , ~ \pm \frac{ 2678 }{ 25 } , ~ \pm \frac{ 2678 }{ 125 } , ~ \pm \frac{ 2678 }{ 625 } , ~ \pm \frac{ 2678 }{ 3125 } , ~ \pm \frac{ 2678 }{ 15625 } ~ $$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{ 13 }{ 25 }) = 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}{ 25 x - 13 } $
$$ \frac{ -15625x^{3}+31250x^{2}-17175x+2678 }{ \color{blue}{ 25x - 13 } } = -625x^{2}+925x-206 $$Polynomial $ -625x^{2}+925x-206 $ can be used to find the remaining roots.
$ \color{blue}{ -625x^{2}+925x-206 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.