College Algebra, Chapter 1, 1.3, Section 1.3, Problem 48

Find all real solutions of $\displaystyle \sqrt{6} x^2 + 2x - \sqrt{\frac{3}{2}} = 0$.


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\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
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