\( c b x ^ { 2 } + 24 x - 30 \)

The expression you've provided is a quadratic polynomial in terms of \( x \): \[ cbx^{2} + 24x - 30 \] To **solve for \( x \)**, you can use the **quadratic formula**. The general form of a quadratic equation is: \[ ax^{2} + bx + c = 0 \] Comparing this with your equation: - \( a = cb \) - \( b = 24 \) - \( c = -30 \) The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] **Plugging in the values:** \[ x = \frac{-24 \pm \sqrt{24^{2} - 4 \cdot cb \cdot (-30)}}{2 \cdot cb} \] **Simplify the discriminant (\( \Delta \)):** \[ \Delta = 24^{2} - 4 \cdot cb \cdot (-30) = 576 + 120cb \] **So, the solutions for \( x \) are:** \[ x = \frac{-24 \pm \sqrt{576 + 120cb}}{2cb} \] **Further Simplification (Optional):** You can factor out a common factor in the numerator and denominator if desired: \[ x = \frac{-24 \pm \sqrt{576 + 120cb}}{2cb} = \frac{-12 \pm \frac{\sqrt{576 + 120cb}}{2}}{cb} \] However, the expression is already in its simplest form unless specific values for \( c \) and \( b \) are provided. --- **Example:** If you have specific values for \( c \) and \( b \), you can substitute them into the equation to find numerical solutions for \( x \). Here's how you'd do it step-by-step with hypothetical values: 1. **Assume** \( c = 2 \) and \( b = 3 \). 2. **Plug** the values into the discriminant: \[ \Delta = 576 + 120 \cdot 2 \cdot 3 = 576 + 720 = 1296 \] 3. **Take** the square root of the discriminant: \[ \sqrt{1296} = 36 \] 4. **Apply** the quadratic formula: \[ x = \frac{-24 \pm 36}{2 \cdot 2 \cdot 3} = \frac{-24 \pm 36}{12} \] 5. **Find** the two possible solutions: - \( x = \frac{-24 + 36}{12} = \frac{12}{12} = 1 \) - \( x = \frac{-24 - 36}{12} = \frac{-60}{12} = -5 \) **Solutions: \( x = 1 \) and \( x = -5 \)** *Note:* Without specific values for \( c \) and \( b \), the solutions will remain in terms of these variables.