$$-500x^3+480x^2+21x-68 = 0$$
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 2 } & x_2 = \dfrac{ 4 }{ 5 } & x_3 = -\dfrac{ 17 }{ 50 } \end{matrix} $$
$ \color{blue}{ -500x^{3}+480x^{2}+21x-68 } $ 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 ( -500 ) are 1 2 4 5 10 20 25 50 100 125 250 500 .The factors of the constant term (-68) are 1 2 4 17 34 68 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 50 } , ~ \pm \frac{ 1 }{ 100 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 1 }{ 250 } , ~ \pm \frac{ 1 }{ 500 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 2 }{ 10 } , ~ \pm \frac{ 2 }{ 20 } , ~ \pm \frac{ 2 }{ 25 } , ~ \pm \frac{ 2 }{ 50 } , ~ \pm \frac{ 2 }{ 100 } , ~ \pm \frac{ 2 }{ 125 } , ~ \pm \frac{ 2 }{ 250 } , ~ \pm \frac{ 2 }{ 500 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 4 }{ 5 } , ~ \pm \frac{ 4 }{ 10 } , ~ \pm \frac{ 4 }{ 20 } , ~ \pm \frac{ 4 }{ 25 } , ~ \pm \frac{ 4 }{ 50 } , ~ \pm \frac{ 4 }{ 100 } , ~ \pm \frac{ 4 }{ 125 } , ~ \pm \frac{ 4 }{ 250 } , ~ \pm \frac{ 4 }{ 500 } , ~ \pm \frac{ 17 }{ 1 } , ~ \pm \frac{ 17 }{ 2 } , ~ \pm \frac{ 17 }{ 4 } , ~ \pm \frac{ 17 }{ 5 } , ~ \pm \frac{ 17 }{ 10 } , ~ \pm \frac{ 17 }{ 20 } , ~ \pm \frac{ 17 }{ 25 } , ~ \pm \frac{ 17 }{ 50 } , ~ \pm \frac{ 17 }{ 100 } , ~ \pm \frac{ 17 }{ 125 } , ~ \pm \frac{ 17 }{ 250 } , ~ \pm \frac{ 17 }{ 500 } , ~ \pm \frac{ 34 }{ 1 } , ~ \pm \frac{ 34 }{ 2 } , ~ \pm \frac{ 34 }{ 4 } , ~ \pm \frac{ 34 }{ 5 } , ~ \pm \frac{ 34 }{ 10 } , ~ \pm \frac{ 34 }{ 20 } , ~ \pm \frac{ 34 }{ 25 } , ~ \pm \frac{ 34 }{ 50 } , ~ \pm \frac{ 34 }{ 100 } , ~ \pm \frac{ 34 }{ 125 } , ~ \pm \frac{ 34 }{ 250 } , ~ \pm \frac{ 34 }{ 500 } , ~ \pm \frac{ 68 }{ 1 } , ~ \pm \frac{ 68 }{ 2 } , ~ \pm \frac{ 68 }{ 4 } , ~ \pm \frac{ 68 }{ 5 } , ~ \pm \frac{ 68 }{ 10 } , ~ \pm \frac{ 68 }{ 20 } , ~ \pm \frac{ 68 }{ 25 } , ~ \pm \frac{ 68 }{ 50 } , ~ \pm \frac{ 68 }{ 100 } , ~ \pm \frac{ 68 }{ 125 } , ~ \pm \frac{ 68 }{ 250 } , ~ \pm \frac{ 68 }{ 500 } ~ $$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 }{ 2 }) = 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}{ 2 x - 1 } $
$$ \frac{ -500x^{3}+480x^{2}+21x-68 }{ \color{blue}{ 2x - 1 } } = -250x^{2}+115x+68 $$Polynomial $ -250x^{2}+115x+68 $ can be used to find the remaining roots.
$ \color{blue}{ -250x^{2}+115x+68 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.