$$-120x^3+82x^2+43x-5 = 0$$
$$ \begin{matrix}x_1 = 1 & x_2 = \dfrac{ 1 }{ 10 } & x_3 = -\dfrac{ 5 }{ 12 } \end{matrix} $$
$ \color{blue}{ -120x^{3}+82x^{2}+43x-5 } $ 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 ( -120 ) are 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120 .The factors of the constant term (-5) are 1 5 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 12 } , ~ \pm \frac{ 1 }{ 15 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 1 }{ 24 } , ~ \pm \frac{ 1 }{ 30 } , ~ \pm \frac{ 1 }{ 40 } , ~ \pm \frac{ 1 }{ 60 } , ~ \pm \frac{ 1 }{ 120 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 3 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 5 }{ 5 } , ~ \pm \frac{ 5 }{ 6 } , ~ \pm \frac{ 5 }{ 8 } , ~ \pm \frac{ 5 }{ 10 } , ~ \pm \frac{ 5 }{ 12 } , ~ \pm \frac{ 5 }{ 15 } , ~ \pm \frac{ 5 }{ 20 } , ~ \pm \frac{ 5 }{ 24 } , ~ \pm \frac{ 5 }{ 30 } , ~ \pm \frac{ 5 }{ 40 } , ~ \pm \frac{ 5 }{ 60 } , ~ \pm \frac{ 5 }{ 120 } ~ $$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(1) = 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 - 1} $
$$ \frac{ -120x^{3}+82x^{2}+43x-5 }{ \color{blue}{ x - 1 } } = -120x^{2}-38x+5 $$Polynomial $ -120x^{2}-38x+5 $ can be used to find the remaining roots.
$ \color{blue}{ -120x^{2}-38x+5 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.