\( \left. \begin{array} { l } { x - 2 } \\ { x ^ { 2 } + x - 6 } \end{array} \right. \)

It looks like you've provided a system of equations or expressions, but it's not entirely clear what you're asking for. The expressions you've written are: 1. \( x - 2 \) 2. \( x^2 + x - 6 \) If you want to solve for \( x \) in the second expression, we can factor it: \[ x^2 + x - 6 = 0 \] To factor this quadratic, we look for two numbers that multiply to \(-6\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(3\) and \(-2\) work: \[ x^2 + 3x - 2x - 6 = 0 \] \[ (x - 2)(x + 3) = 0 \] Setting each factor to zero gives us the solutions: 1. \( x - 2 = 0 \) → \( x = 2 \) 2. \( x + 3 = 0 \) → \( x = -3 \) So the solutions to the equation \( x^2 + x - 6 = 0 \) are \( x = 2 \) and \( x = -3 \). If you meant something else or have a specific question regarding these expressions, please provide more details!