$$(6x+5)(3x^2+14x-5) = 0$$

$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = -5 & x_3 = -\dfrac{ 5 }{ 6 } \end{matrix} $$

$ \color{blue}{ 18x^{3}+99x^{2}+40x-25 } $ 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 ( 18 ) are 1 2 3 6 9 18 .The factors of the constant term (-25) are 1 5 25 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 18 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 3 } , ~ \pm \frac{ 5 }{ 6 } , ~ \pm \frac{ 5 }{ 9 } , ~ \pm \frac{ 5 }{ 18 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 25 }{ 2 } , ~ \pm \frac{ 25 }{ 3 } , ~ \pm \frac{ 25 }{ 6 } , ~ \pm \frac{ 25 }{ 9 } , ~ \pm \frac{ 25 }{ 18 } ~ $$

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 }{ 3 }) = 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}{ 3 x - 1 } $

$$ \frac{ 18x^{3}+99x^{2}+40x-25 }{ \color{blue}{ 3x - 1 } } = 6x^{2}+35x+25 $$

Polynomial $ 6x^{2}+35x+25 $ can be used to find the remaining roots.

$ \color{blue}{ 6x^{2}+35x+25 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

Polynomial Equations Solver