$$(x^2+6x)(x^2+6x+13) = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -6 & x_3 = -3+2i \\[1 em] x_4 = -3-2i & \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ x^{4}+12x^{3}+49x^{2}+78x = 0 } $, first we need to factor our $ x $.

$$ x^{4}+12x^{3}+49x^{2}+78x = x \left( x^{3}+12x^{2}+49x+78 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ x^{3}+12x^{2}+49x+78 = 0$.

$ \color{blue}{ x^{3}+12x^{2}+49x+78 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (78) are 1 2 3 6 13 26 39 78 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 13 }{ 1 } , ~ \pm \frac{ 26 }{ 1 } , ~ \pm \frac{ 39 }{ 1 } , ~ \pm \frac{ 78 }{ 1 } ~ $$

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(-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}{x + 6} $

$$ \frac{ x^{3}+12x^{2}+49x+78 }{ \color{blue}{ x + 6 } } = x^{2}+6x+13 $$

Polynomial $ x^{2}+6x+13 $ can be used to find the remaining roots.

$ \color{blue}{ x^{2}+6x+13 } $ 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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