Precalculus (6th Edition) Blitzer

The solution of the equation $3{{x}^{2}}=8x-7$ in standard form is $\left\{ \frac{4}{3}+\frac{\sqrt{5}}{3}i,\frac{4}{3}-\frac{\sqrt{5}}{3}i \right\}$.
Consider the equation,$3{{x}^{2}}=8x-7$ Rearrange the equation. \begin{align} & 3{{x}^{2}}-\left( 8x-7 \right)=0 \\ & 3{{x}^{2}}-8x+7=0 \end{align} Compare the equation $3{{x}^{2}}-8x+7=0$ with $a{{x}^{2}}+bx+c$. \begin{align} & a=3 \\ & b=-8 \\ & c=7 \end{align} Substitute $a=3$, $b=-8$ and $c=7$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. \begin{align} & x=\frac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 3 \right)\left( 7 \right)}}{2\left( 3 \right)} \\ & =\frac{8\pm \sqrt{64-84}}{6} \\ & =\frac{8\pm \sqrt{-20}}{6} \end{align} Use the property $\sqrt{-b}=i\sqrt{b}$. \begin{align} & x=\frac{8\pm i\sqrt{20}}{6} \\ & =\frac{8\pm i\sqrt{4\cdot 5}}{6} \\ & =\frac{8\pm 2i\sqrt{5}}{6} \end{align} Express the complex number in the standard form. \begin{align} & x=\frac{8}{6}\pm \frac{2\sqrt{5}}{6}i \\ & =\frac{4}{3}\pm \frac{\sqrt{5}}{3}i \end{align} Therefore, the solution of the equation $3{{x}^{2}}=8x-7$ in standard form is $\left\{ \frac{4}{3}+\frac{\sqrt{5}}{3}i,\frac{4}{3}-\frac{\sqrt{5}}{3}i \right\}$.