$$(3e^2-5e+7)(6e+1) = 0$$

$$ \begin{matrix}e_1 = -\dfrac{ 1 }{ 6 } & e_2 = \dfrac{ 5 }{ 6 }+\dfrac{\sqrt{ 59 }}{ 6 }i & e_3 = \dfrac{ 5 }{ 6 }-\dfrac{\sqrt{ 59 }}{ 6 }i \end{matrix} $$

$ \color{blue}{ 18x^{3}-27x^{2}+37x+7 } $ 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 (7) are 1 7 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 18 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 2 } , ~ \pm \frac{ 7 }{ 3 } , ~ \pm \frac{ 7 }{ 6 } , ~ \pm \frac{ 7 }{ 9 } , ~ \pm \frac{ 7 }{ 18 } ~ $$

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 }{ 6 }) = 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}{ 6 x + 1 } $

$$ \frac{ 18x^{3}-27x^{2}+37x+7 }{ \color{blue}{ 6x + 1 } } = 3x^{2}-5x+7 $$

Polynomial $ 3x^{2}-5x+7 $ can be used to find the remaining roots.

$ \color{blue}{ 3x^{2}-5x+7 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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