Solve the equation by completing the square and applying the square root property, \[ 3 x^{2}=4=2 x \]

To solve the equation \( 3x^2 = 4 - 2x \), start by rearranging it to the standard form. Rewrite it as \( 3x^2 + 2x - 4 = 0 \). Next, we need to complete the square. 1. Divide everything by 3 to simplify: \[ x^2 + \frac{2}{3}x - \frac{4}{3} = 0 \] 2. Move \(-\frac{4}{3}\) to the other side: \[ x^2 + \frac{2}{3}x = \frac{4}{3} \] 3. To complete the square, take half of the coefficient of \(x\) (which is \(\frac{2}{3}\)), square it, and add it to both sides. Half of \(\frac{2}{3}\) is \(\frac{1}{3}\), and squaring it gives \(\frac{1}{9}\): \[ x^2 + \frac{2}{3}x + \frac{1}{9} = \frac{4}{3} + \frac{1}{9} \] 4. Find a common denominator for the right side: \[ \frac{4}{3} = \frac{12}{9} \Rightarrow \frac{12}{9} + \frac{1}{9} = \frac{13}{9} \] 5. Now we have: \[ \left(x + \frac{1}{3}\right)^2 = \frac{13}{9} \] 6. Take the square root of both sides: \[ x + \frac{1}{3} = \pm \frac{\sqrt{13}}{3} \] 7. Finally, isolate \(x\): \[ x = -\frac{1}{3} \pm \frac{\sqrt{13}}{3} \] So the solutions are: \[ x = \frac{-1 + \sqrt{13}}{3} \quad \text{and} \quad x = \frac{-1 - \sqrt{13}}{3} \]