College Algebra, Chapter 1, 1.3, Section 1.3, Problem 38

Find all real solutions of $8x^2 - 6x - 9 = 0$.


$
\begin{equation}
\begin{aligned}

8x^2 - 6x - 9 =& 0
&& \text{Given}
\\
\\
8x^2 - 6x =& 9
&& \text{Add 9}
\\
\\
x^2 - \frac{6}{8} x =& \frac{9}{8}
&& \text{Divide both sides of the equation by 8 to make the coefficient of $x^2$ equal to 1}
\\
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x^2 - \frac{6}{8} x + \frac{9}{64} =& \frac{9}{8} + \frac{9}{64}
&& \text{Complete the the square: add } \left( \frac{\displaystyle \frac{-6}{8}}{2} \right)^2 = \frac{9}{64}
\\
\\
\left( x - \frac{3}{8} \right)^2 =& \frac{81}{64}
&& \text{Perfect square, get the LCD of the right side}
\\
\\
x - \frac{3}{8} =& \pm \sqrt{\frac{81}{64}}
&& \text{Take the square root}
\\
\\
x =& \frac{3}{8} \pm \frac{9}{8}
&& \text{Add } \frac{3}{8}
\\
\\
x =& \frac{3}{2} \text{ and } x = \frac{-3}{4}
&& \text{Solve for } x

\end{aligned}
\end{equation}
$