Task B-2.1. Determine all values of the real parameter $a$ for which the sum of the squares of the solutions of the quadratic equation $x^{2}+2 a x+a-3=0$ is greater than 6.
Solution. From Vieta's formulas, it follows that

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=4 a^{2}-2 a+6
$$

It must hold that

$$
\begin{array}{r}
4 a^{2}-2 a+6>6 \\
4 a^{2}-2 a>0 \\
2 a(2 a-1)>0
\end{array}
$$

The sought values of the parameter $a$ are the solutions to this inequality, that is

$$
a \in\langle-\infty, 0\rangle \cup\left\langle\frac{1}{2}, \infty\right\rangle
$$