Precalculus (6th Edition) Blitzer

The solution of the equation $3{{x}^{2}}=4x-6$ in standard form is $\left\{ \frac{2}{3}+\frac{\sqrt{14}}{3}i,\frac{2}{3}-\frac{\sqrt{14}}{3}i \right\}$.
Consider the equation,$3{{x}^{2}}=4x-6$ Rearrange the equation. \begin{align} & 3{{x}^{2}}-\left( 4x-6 \right)=0 \\ & 3{{x}^{2}}-4x+6=0 \end{align} Compare the equation $3{{x}^{2}}-4x+6=0$ with $a{{x}^{2}}+bx+c$. \begin{align} & a=3 \\ & b=-4 \\ & c=6 \end{align} Substitute $a=3$, $b=-4$ and $c=6$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. \begin{align} & x=\frac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 3 \right)\left( 6 \right)}}{2\left( 3 \right)} \\ & =\frac{4\pm \sqrt{16-72}}{6} \\ & =\frac{4\pm \sqrt{-56}}{6} \end{align} Use the property $\sqrt{-b}=i\sqrt{b}$. \begin{align} & x=\frac{4\pm i\sqrt{56}}{6} \\ & =\frac{4\pm i\sqrt{4\cdot 14}}{6} \\ & =\frac{4\pm 2i\sqrt{14}}{6} \end{align} Express the complex number in the standard form. \begin{align} & x=\frac{4}{6}\pm \frac{2\sqrt{14}}{6}i \\ & =\frac{2}{3}\pm \frac{\sqrt{14}}{3}i \end{align} Therefore, the solution of the equation $3{{x}^{2}}=4x-6$ in standard form is $\left\{ \frac{2}{3}+\frac{\sqrt{14}}{3}i,\frac{2}{3}-\frac{\sqrt{14}}{3}i \right\}$.