How could I know that 4x^4 + 3^2 - 3 has two roots?
You don't. It's not true. Or are you asking about real roots?
Is that really the problem? $\displaystyle 4x^4+ 3^2- 3= 4x^4+ 9- 3= 4x^4+ 6= 0$ so $\displaystyle x^4= -6/4$ so that there are four roots, the four fourth roots of -3/2, none of which are real.
If the equation is $\displaystyle 4x^4+ x^2- 3= 0$, we can let $\displaystyle y= x^2$ so we are dealing with the quadratic equation $\displaystyle 4y^2+ y- 3= 0$. By the quadratic formula,
$\displaystyle y= \frac{-1\pm\sqrt{1+ 48}}{8}= \frac{-1\pm 7}{8}$
which has the two roots y= 6/8= 3/4 and y= -8/8= -1.
Now, If $\displaystyle y= x^2= 3/4$, then $\displaystyle x= \pm\sqrt{3}/2$, two real roots. If $\displaystyle y= x^2= -1$, then $\displaystyle x= \pm i$, two imaginary roots.
The crucial point was that, because this had only even factors, we could reduce to a second order equation. Positive roots of that equation give real roots of the original equation.