College Algebra, Chapter 1, 1.4, Section 1.4, Problem 70

Find all solutions of the equation $4x^2 - 16x + 19 = 0$ and express them in the form $a + bi$.


$\begin{equation} \begin{aligned} 4x^2 - 16x + 19 =& 0 && \text{Given} \\ \\ 4x^2 - 16x =& -19 && \text{Subtract } 19 \\ \\ x^2 - 4x =& \frac{-19}{4} && \text{Divide both sides of the equation by } 4 \\ \\ x^2 - 4x + 4 =& \frac{-19}{4} + 4 && \text{Complete the square: add } \left( \frac{-4}{2} \right)^2 = 4 \\ \\ (x - 2)^2 =& \frac{-3}{4} && \text{Perfect square} \\ \\ x - 2 =& \pm \sqrt{\frac{-3}{4}} && \text{Take the square root} \\ \\ x - 2 =& \pm \sqrt{\frac{3i^2}{4}} && \text{Recall that } i^2 = -1 \\ \\ x =& 2 \pm \frac{\sqrt{3}}{2} i && \text{Add } 2 \text{ and simplify} \\ \\ \left( x - \left( 2 + \frac{\sqrt{3}}{2} i \right) \right)& \left( x - \left( 2 - \frac{\sqrt{3}}{2} i \right) \right) = 0 && \end{aligned} \end{equation}$