Let's determine the real roots of the following equation:

$$
2 x^{4}+4 x^{3}+3 x^{2}+x-1=0
$$
I. solution: Dividing both sides by 2

$$
x^{4}+2 x^{3}+\frac{3}{2} x^{2}+\frac{1}{2} x-\frac{1}{2}=0
$$

In this case, the necessary and sufficient condition for the quartic equation $x^{4}+a x^{3}+b x^{2}+c x+d=0$ to be reducible to a quadratic equation (as stated in the 656th problem of the 2nd issue of the 11th volume of Középiskolai Mat. Lapok, p. 57) is:

$$
a^{3}-4 a b+8 c=8-12+4=0
$$

According to the information provided there, the transformation $x=z-\frac{a}{4}=z-\frac{1}{2}$ will reduce our equation to a quadratic one:

$$
\begin{gathered}
\left(z-\frac{1}{2}\right)^{4}+2\left(z-\frac{1}{2}\right)^{3}+\frac{3}{2}\left(z-\frac{1}{2}\right)^{2}+\frac{1}{2}\left(z-\frac{1}{2}\right)-\frac{1}{2}=z^{4}-\frac{9}{16}=0 \\
z^{4}=\frac{9}{16}
\end{gathered}
$$

From this, the real roots are

$$
z_{1}=\frac{\sqrt{3}}{2} \quad \text { and } \quad z_{2}=-\frac{\sqrt{3}}{2}
$$

and thus

$$
x_{1}=z_{1}-\frac{1}{2}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \quad \text { and } \quad x_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} .
$$

This gives us the real roots of the given equation.

András Sárközy (Gyöngyös, Vak Bottyán g. III. o. t.)

II. solution: Rearrange the equation:

$$
2 x^{4}+4 x^{3}+2 x^{2}+x^{2}+x-1=0
$$

After factoring out:

$$
\begin{gathered}
2 x^{2}\left(x^{2}+2 x-1\right)+x(x+1)-1=0 \\
2[x(x+1)]^{2}+x(x+1)-1=0
\end{gathered}
$$

Substituting $x(x+1)=y$:

$$
2 y^{2}+y-1=0
$$

from which

$$
y_{1}=\frac{1}{2}, \quad y_{2}=-1
$$

Substituting $y$ back into $x(x+1)=y$ gives us two equations for $x$:

$$
x^{2}+x-\frac{1}{2}=0 \quad \text { and } \quad x^{2}+x+1=0
$$

The discriminant of the second equation is negative, so it has no real roots, and the roots of the first equation give the values found in the first solution.

Katalin Fanta (Szombathely, Kanizsai D. lg. III. o. t.)

III. solution: Add $x^{2}+x$ to both sides of the equation and then subtract the same amount, then rearrange the equation:

$$
\begin{gathered}
2 x^{4}+2 x^{3}+2 x^{2}+2 x^{3}+2 x^{2}+2 x-x^{2}-x-1=0 \\
2 x^{2}\left(x^{2}+x+1\right)+2 x\left(x^{2}+x+1\right)-\left(x^{2}+x+1\right)=0
\end{gathered}
$$

Factoring out:

$$
\left(x^{2}+x+1\right)\left(2 x^{2}+2 x-1\right)=0
$$

and thus either

$$
x^{2}+x+1=0
$$

or

$$
2 x^{2}+2 x-1=0
$$

We obtained the same equations as at the end of the second solution, and we get the same values for the real roots.