3(х-2)(х+4)=2х+

To solve the equation \( 3(x - 2)(x + 4) = 2x + \) (it seems that the right side may be incomplete), I will first assume you want to solve the left side as an expression and set it equal to something, or if the right side was meant to be \( 2x \), I will present the steps for that. Let's first expand the left side: 1. Expand \( (x - 2)(x + 4) \): \[ (x - 2)(x + 4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8 \] 2. Now multiply by 3: \[ 3(x^2 + 2x - 8) = 3x^2 + 6x - 24 \] Thus, we have: \[ 3(x - 2)(x + 4) = 3x^2 + 6x - 24 \] Now, let's consider setting this equal to \( 2x \) which seems to be part of your equation: \[ 3x^2 + 6x - 24 = 2x \] 3. Rearranging the equation gives: \[ 3x^2 + 6x - 2x - 24 = 0 \] \[ 3x^2 + 4x - 24 = 0 \] 4. To solve this quadratic equation, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = 4, c = -24 \). 5. First, calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4(3)(-24) = 16 + 288 = 304 \] 6. Now apply the quadratic formula: \[ x = \frac{-4 \pm \sqrt{304}}{2 \cdot 3} \] \[ x = \frac{-4 \pm 2\sqrt{76}}{6} \] \[ x = \frac{-2 \pm \sqrt{76}}{3} \] Now, simplifying \( \sqrt{76} \): \[ \sqrt{76} = 2\sqrt{19} \] Thus, we have: \[ x = \frac{-2 \pm 2\sqrt{19}}{3} = \frac{-1 \pm \sqrt{19}}{3} \] So the solutions for the equation \( 3(x - 2)(x + 4) = 2x \) are: \[ x = \frac{-1 + \sqrt{19}}{3}, \quad x = \frac{-1 - \sqrt{19}}{3} \] If your question was different or if the right side of your equation was meant to be something else, please let me know!