Solving $10x^4-13x^2+4=0$

Hint:

Let $y = x^2$. Then you're solving $10y^2 - 13y + 4 = 0$.

You have a quartic but it is simply a quadratic in disguise of a quartic so you may use the quadratic formula: $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ For $$x^4 -13x^2+4=0$$ we have $$x^2=\frac{13\pm \sqrt{13^2-4 \times 10 \times 4}}{2\times 10}=\frac{13\pm 3}{20}$$

So $$x^2=\frac12, \frac45 \implies x=\pm {\frac{1}{\sqrt2}},\space\pm {\frac{2}{\sqrt5}}$$

You can directly factorize as $10x^4-5x^2-8x^2+4=0$ and then proceed normally by taking $x^2$ common from first two terms and solve like a quadratic to get $4$ roots. ie $\pm \sqrt{0.8}, \pm \sqrt{0.5} = x$.