College Algebra, Chapter 1, 1.3, Section 1.3, Problem 42

Find all real solutions of $\displaystyle 2y^2 - y - \frac{1}{2} = 0$.


$
\begin{equation}
\begin{aligned}

2y^2 - y - \frac{1}{2} =& 0
&& \text{Given}
\\
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2y^2 - y =& \frac{1}{2}
&& \text{Add } \frac{1}{2}
\\
\\
y^2 - \frac{y}{2} =& \frac{1}{4}
&& \text{Divide both sides by 2 to make the coefficient of $x^2$ equal to 1}
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y^2 - \frac{y}{2} + \frac{1}{16} =& \frac{1}{4} + \frac{1}{16}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{-1}{2}}{2} \right)^2 = \frac{1}{16}
\\
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\left( y - \frac{1}{4} \right)^2 =& \frac{5}{16}
&& \text{Perfect square, get the LCD of the right side of the equation}
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y - \frac{1}{4} =& \pm \sqrt{\frac{5}{16}}
&& \text{Take the square root}
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y =& \frac{1}{4} \pm \sqrt{\frac{5}{16}}
&& \text{Add } \frac{1}{4}
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y =& \frac{1}{4} + \frac{\sqrt{5}}{4} \text{ and } y = \frac{1}{4} - \frac{\sqrt{5}}{4}
&& \text{Solve for } y
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y =& \frac{1 + \sqrt{5}}{4} \text{ and } y = \frac{1 - \sqrt{5}}{4}
&& \text{Simplify}

\end{aligned}
\end{equation}
$