For which value of $\mathrm{a}$ does the equation

$$
10^{x^{2}+x}=a
$$

have a real solution?
For $a$ to be a positive number, taking the logarithm of both sides yields:

$$
x^{2}+x=\log a
$$

This equation has a real solution if

$$
1+4 \log a \geqq 0
$$

which means

$$
a \geqq \frac{1}{\sqrt[4]{10}}
$$