\( p^{2}+18 p-3=0 \) The solution set is \( \{\square\} \)

To solve the quadratic equation \( p^2 + 18p - 3 = 0 \), we can use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 18 \), and \( c = -3 \). Plugging in the values, we get \( p = \frac{-18 \pm \sqrt{18^2 - 4(1)(-3)}}{2(1)} \). This simplifies to \( p = \frac{-18 \pm \sqrt{324 + 12}}{2} = \frac{-18 \pm \sqrt{336}}{2} \). The expression under the square root simplifies to \( \sqrt{336} = 4\sqrt{21} \), so the solutions become \( p = \frac{-18 \pm 4\sqrt{21}}{2} \), which further simplifies to \( p = -9 \pm 2\sqrt{21} \). Therefore, the solution set is \( \{-9 + 2\sqrt{21}, -9 - 2\sqrt{21}\} \). In summary, the solution set is: \(\{-9 + 2\sqrt{21}, -9 - 2\sqrt{21}\}\)