$$x^2+\frac{4}{5x^2}-3 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = - \dfrac{\sqrt{ 15 }}{ 3 } & x_3 = \dfrac{\sqrt{ 15 }}{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} x^2+\frac{4}{5x^2}-3 &= 0&& \text{multiply ALL terms by } \color{blue}{ 5x^2 }. \\[1 em]5x^2x^2+5x^2\cdot\frac{4}{5x^2}-5x^2\cdot3 &= 5x^2\cdot0&& \text{cancel out the denominators} \\[1 em]5x^4+4x^4-15x^2 &= 0&& \text{simplify left side} \\[1 em]9x^4-15x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 9x^{4}-15x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 9x^{4}-15x^{2} = x^2 \left( 9x^{2}-15 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 9x^{2}-15 = 0$.
$ 9x^{2}-15 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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