\[\begin{array}{l} {x^4} - 22{x^2} - 5x + 2 = \left( {{x^2} + ax + 1} \right) \cdot \left( {{x^2} + bx + 2} \right) = \\ {x^4} + b{x^3} + 2{x^2} + a{x^3} + ab{x^2} + 2ax + {x^2} + bx + 2 = \\ {x^4} + \left( {a + b} \right){x^3} + \left( {3 + ab} \right){x^2} + \left( {2a + b} \right)x + 2\\ \left\{ \begin{array}{l} a + b = 0\\ 2a + b = - 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = - a\\ 2a - a = - 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = 5\\ a = - 5 \end{array} \right. \Rightarrow \\ {x^4} - 22{x^2} - 5x + 2 = \left( {{x^2} - 5x + 1} \right) \cdot \left( {{x^2} + 5x + 2} \right)\\ {x^4} - 22{x^2} - 5x + 2 = 0\\ \left( {{x^2} - 5x + 1} \right) \cdot \left( {{x^2} + 5x + 2} \right) = 0\\ \left[ \begin{array}{l} {x^2} - 5x + 1 = 0\\ {x^2} + 5x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x_{1,2}} = \frac{{5 \pm \sqrt {21} }}{2}\\ {x_{3,4}} = \frac{{ - 5 \pm \sqrt {17} }}{2} \end{array} \right. \end{array}\]

\[x \in \left\{ {\frac{{5 \pm \sqrt {21} }}{2};\frac{{ - 5 \pm \sqrt {17} }}{2}} \right\}\]