## Task 2 - 291222

Determine all real numbers $m$ that satisfy the condition that for every real number $x$, the following inequality (1) holds:

$$
x^{2}+(m+2) x+8 m+1>0
$$
By completing the square, the inequality can be represented as follows:

$$
\left(x+\frac{m+2}{2}\right)^{2}+8 m+1-\left(\frac{m+2}{2}\right)^{2}>0
$$

This inequality is always satisfied for all real $x$ if

$$
\begin{gathered}
8 m+1-\left(\frac{m+2}{2}\right)^{2}>0 \\
32 m+4-m^{2}-4 m-4>0 \\
m^{2}-28 m<0
\end{gathered}
$$

The inequality is satisfied for $0<m<28$.