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}
$