College Algebra, Chapter 1, 1.4, Section 1.4, Problem 66

Find all solutions of the equation $2x^2 + 3 = 2x$ and express them in the form $a + bi$.


$\begin{equation} \begin{aligned} 2x^2 + 3 =& 2x && \text{Given} \\ \\ 2x^2 - 2x + 3 =& 0 && \text{Subtract } 2x \text{ to both sides of the equation} \\ \\ x^2 - x + \frac{3}{2} =& 0 && \text{Subtract } \frac{3}{2} \\ \\ x^2 - x + \frac{1}{4} =& \frac{-3}{2} + \frac{1}{4} && \text{Complete the square: add } \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \\ \\ \left( x - \frac{1}{2} \right)^2 =& \frac{-5}{4} && \text{Perfect square} \\ \\ x - \frac{1}{2} =& \pm \sqrt{\frac{-5}{4}} && \text{Take the square root} \\ \\ x - \frac{1}{2} =& \pm \sqrt{\frac{5i^2}{4}} && \text{Recall that } i^2 = -1 \\ \\ x =& \frac{1}{2} \pm \frac{\sqrt{5}}{2} i && \text{Add } \frac{1}{2} \text{ and simplify} \\ \\ \left( x - \left( \frac{1 + \sqrt{5}}{2} \right) \right) & \left( x - \left( \frac{1 - \sqrt{5}}{2} \right) \right) = 0 && \end{aligned} \end{equation}$