for \( x \) in each of the follo \[ \begin{array}{l}x(2 x+1)=0 \\ 5 x^{2}+2 x-6=0 \\ 2 x^{2}-2 \geq 3 x\end{array} \]

To solve for \( x \) in the given equations, let's break them down step by step. 1. For \( x(2x + 1) = 0 \): This equation can be solved by setting each factor equal to zero. - \( x = 0 \) - \( 2x + 1 = 0 \) → \( 2x = -1 \) → \( x = -\frac{1}{2} \) Thus, the solutions are \( x = 0 \) and \( x = -\frac{1}{2} \). 2. For \( 5x^{2} + 2x - 6 = 0 \): Here, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5 \), \( b = 2 \), and \( c = -6 \). - Discriminant: \( b^2 - 4ac = 2^2 - 4 \cdot 5 \cdot (-6) = 4 + 120 = 124 \) - Roots: \[ x = \frac{-2 \pm \sqrt{124}}{2 \cdot 5} = \frac{-2 \pm 2\sqrt{31}}{10} = \frac{-1 \pm \sqrt{31}}{5} \] Therefore, the solutions are: \[ x = \frac{-1 + \sqrt{31}}{5} \quad \text{and} \quad x = \frac{-1 - \sqrt{31}}{5} \] 3. For \( 2x^{2} - 2 \geq 3x \): Rearranging gives us \( 2x^{2} - 3x - 2 \geq 0 \). We can factor or use the quadratic formula. - Using the quadratic formula \( a = 2, b = -3, c = -2 \): \[ x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \] - Roots: \[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-2}{4} = -\frac{1}{2} \] Now, we can test intervals around \( x = 2 \) and \( x = -\frac{1}{2} \) to identify valid ranges. The inequality holds for \( x \leq -\frac{1}{2} \) or \( x \geq 2 \). In summary: - From the first equation: \( x = 0 \) and \( x = -\frac{1}{2} \) - From the second equation: \( x = \frac{-1 + \sqrt{31}}{5} \) and \( x = \frac{-1 - \sqrt{31}}{5} \) - From the third inequality: \( x \leq -\frac{1}{2} \) or \( x \geq 2 \)