Solve the following equation:

$$
\left|x^{2}+x-\frac{21}{16}\right|=\left|x^{2}-1\right| \text {. }
$$
The absolute value of two numbers or expressions can be equal in two ways: 1. they have the same sign, meaning their difference is 0; or 2. they have opposite signs, meaning their sum is 0. (The two possibilities coincide if the value of both expressions is 0.) According to this, (1) holds if

$$
\begin{aligned}
& \left(x^{2}+x-\frac{21}{16}\right)-\left(x^{2}-1\right)=x-\frac{5}{16}=0, \text { and if } \\
& \left(x^{2}+x-\frac{21}{16}\right)+\left(x^{2}-1\right)=2 x^{2}+x-\frac{37}{16}=0
\end{aligned}
$$

From the first, $x_{1}=5 / 16=0.3125$, from the second, $x_{2}=-1 / 4+\sqrt{39 / 32} \approx 0.854, x_{3}=-1 / 4-\sqrt{39 / 32} \approx-1.354$.

János Gáspár (Nagybátony, B. Bartók Secondary School, 2nd grade)