Answer
The solution of the equation $3{{x}^{2}}=4x-6$ in standard form is $\left\{ \frac{2}{3}+\frac{\sqrt{14}}{3}i,\frac{2}{3}-\frac{\sqrt{14}}{3}i \right\}$.
Work Step by Step
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 $\left\{ \frac{2}{3}+\frac{\sqrt{14}}{3}i,\frac{2}{3}-\frac{\sqrt{14}}{3}i \right\}$.
