2. Given that $a$ is a real number, and makes the quadratic equation in $x$, $x^{2}+a^{2} x+a=0$, have real roots. Then the maximum value that $x$ can take is $\qquad$
2. $\frac{\sqrt[3]{2}}{2}$.

Given that $a$ is a real number, when $a \neq 0$, the quadratic equation in $a$, $a^{2} x+a+x^{2}=0$, has real roots. Therefore, $\Delta=1-4 x^{3} \geqslant 0$. Hence, $x \leqslant \frac{\sqrt[3]{2}}{2}$. When $a=0$, $x=0$.