$$3(x^2-49)(36x^2-49) = 0$$
$$ \begin{matrix}x_1 = 7 & x_2 = -7 & x_3 = \dfrac{ 7 }{ 6 } \\[1 em] x_4 = -\dfrac{ 7 }{ 6 } & \\[1 em] \end{matrix} $$
$ \color{blue}{ 108x^{4}-5439x^{2}+7203 } $ 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 ( 108 ) are 1 2 3 4 6 9 12 18 27 36 54 108 .The factors of the constant term (7203) are 1 3 7 21 49 147 343 1029 2401 7203 . 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 }{ 6 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 12 } , ~ \pm \frac{ 1 }{ 18 } , ~ \pm \frac{ 1 }{ 27 } , ~ \pm \frac{ 1 }{ 36 } , ~ \pm \frac{ 1 }{ 54 } , ~ \pm \frac{ 1 }{ 108 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 3 }{ 9 } , ~ \pm \frac{ 3 }{ 12 } , ~ \pm \frac{ 3 }{ 18 } , ~ \pm \frac{ 3 }{ 27 } , ~ \pm \frac{ 3 }{ 36 } , ~ \pm \frac{ 3 }{ 54 } , ~ \pm \frac{ 3 }{ 108 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 2 } , ~ \pm \frac{ 7 }{ 3 } , ~ \pm \frac{ 7 }{ 4 } , ~ \pm \frac{ 7 }{ 6 } , ~ \pm \frac{ 7 }{ 9 } , ~ \pm \frac{ 7 }{ 12 } , ~ \pm \frac{ 7 }{ 18 } , ~ \pm \frac{ 7 }{ 27 } , ~ \pm \frac{ 7 }{ 36 } , ~ \pm \frac{ 7 }{ 54 } , ~ \pm \frac{ 7 }{ 108 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 21 }{ 2 } , ~ \pm \frac{ 21 }{ 3 } , ~ \pm \frac{ 21 }{ 4 } , ~ \pm \frac{ 21 }{ 6 } , ~ \pm \frac{ 21 }{ 9 } , ~ \pm \frac{ 21 }{ 12 } , ~ \pm \frac{ 21 }{ 18 } , ~ \pm \frac{ 21 }{ 27 } , ~ \pm \frac{ 21 }{ 36 } , ~ \pm \frac{ 21 }{ 54 } , ~ \pm \frac{ 21 }{ 108 } , ~ \pm \frac{ 49 }{ 1 } , ~ \pm \frac{ 49 }{ 2 } , ~ \pm \frac{ 49 }{ 3 } , ~ \pm \frac{ 49 }{ 4 } , ~ \pm \frac{ 49 }{ 6 } , ~ \pm \frac{ 49 }{ 9 } , ~ \pm \frac{ 49 }{ 12 } , ~ \pm \frac{ 49 }{ 18 } , ~ \pm \frac{ 49 }{ 27 } , ~ \pm \frac{ 49 }{ 36 } , ~ \pm \frac{ 49 }{ 54 } , ~ \pm \frac{ 49 }{ 108 } , ~ \pm \frac{ 147 }{ 1 } , ~ \pm \frac{ 147 }{ 2 } , ~ \pm \frac{ 147 }{ 3 } , ~ \pm \frac{ 147 }{ 4 } , ~ \pm \frac{ 147 }{ 6 } , ~ \pm \frac{ 147 }{ 9 } , ~ \pm \frac{ 147 }{ 12 } , ~ \pm \frac{ 147 }{ 18 } , ~ \pm \frac{ 147 }{ 27 } , ~ \pm \frac{ 147 }{ 36 } , ~ \pm \frac{ 147 }{ 54 } , ~ \pm \frac{ 147 }{ 108 } , ~ \pm \frac{ 343 }{ 1 } , ~ \pm \frac{ 343 }{ 2 } , ~ \pm \frac{ 343 }{ 3 } , ~ \pm \frac{ 343 }{ 4 } , ~ \pm \frac{ 343 }{ 6 } , ~ \pm \frac{ 343 }{ 9 } , ~ \pm \frac{ 343 }{ 12 } , ~ \pm \frac{ 343 }{ 18 } , ~ \pm \frac{ 343 }{ 27 } , ~ \pm \frac{ 343 }{ 36 } , ~ \pm \frac{ 343 }{ 54 } , ~ \pm \frac{ 343 }{ 108 } , ~ \pm \frac{ 1029 }{ 1 } , ~ \pm \frac{ 1029 }{ 2 } , ~ \pm \frac{ 1029 }{ 3 } , ~ \pm \frac{ 1029 }{ 4 } , ~ \pm \frac{ 1029 }{ 6 } , ~ \pm \frac{ 1029 }{ 9 } , ~ \pm \frac{ 1029 }{ 12 } , ~ \pm \frac{ 1029 }{ 18 } , ~ \pm \frac{ 1029 }{ 27 } , ~ \pm \frac{ 1029 }{ 36 } , ~ \pm \frac{ 1029 }{ 54 } , ~ \pm \frac{ 1029 }{ 108 } , ~ \pm \frac{ 2401 }{ 1 } , ~ \pm \frac{ 2401 }{ 2 } , ~ \pm \frac{ 2401 }{ 3 } , ~ \pm \frac{ 2401 }{ 4 } , ~ \pm \frac{ 2401 }{ 6 } , ~ \pm \frac{ 2401 }{ 9 } , ~ \pm \frac{ 2401 }{ 12 } , ~ \pm \frac{ 2401 }{ 18 } , ~ \pm \frac{ 2401 }{ 27 } , ~ \pm \frac{ 2401 }{ 36 } , ~ \pm \frac{ 2401 }{ 54 } , ~ \pm \frac{ 2401 }{ 108 } , ~ \pm \frac{ 7203 }{ 1 } , ~ \pm \frac{ 7203 }{ 2 } , ~ \pm \frac{ 7203 }{ 3 } , ~ \pm \frac{ 7203 }{ 4 } , ~ \pm \frac{ 7203 }{ 6 } , ~ \pm \frac{ 7203 }{ 9 } , ~ \pm \frac{ 7203 }{ 12 } , ~ \pm \frac{ 7203 }{ 18 } , ~ \pm \frac{ 7203 }{ 27 } , ~ \pm \frac{ 7203 }{ 36 } , ~ \pm \frac{ 7203 }{ 54 } , ~ \pm \frac{ 7203 }{ 108 } ~ $$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(7) = 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 - 7} $
$$ \frac{ 108x^{4}-5439x^{2}+7203 }{ \color{blue}{ x - 7 } } = 108x^{3}+756x^{2}-147x-1029 $$Polynomial $ 108x^{3}+756x^{2}-147x-1029 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 108x^{3}+756x^{2}-147x-1029 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.