$$2x^4-x^3-18x^2+14x+15 = 0$$

$$ \begin{matrix}x_1 = -3 & x_2 = \dfrac{ 5 }{ 2 } & x_3 = \dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 5 }}{ 2 } \\[1 em] x_4 = \dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 5 }}{ 2 } & \\[1 em] \end{matrix} $$

$ \color{blue}{ 2x^{4}-x^{3}-18x^{2}+14x+15 } $ is a polynomial of degree 4. 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 ( 2 ) are 1 2 .The factors of the constant term (15) are 1 3 5 15 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } ~ $$

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(-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}{x + 3} $

$$ \frac{ 2x^{4}-x^{3}-18x^{2}+14x+15 }{ \color{blue}{ x + 3 } } = 2x^{3}-7x^{2}+3x+5 $$

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

Use the same procedure to find roots of $ 2x^{3}-7x^{2}+3x+5 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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