Answer
The solution of the equation $3{{x}^{2}}=8x-7$ in standard form is $\left\{ \frac{4}{3}+\frac{\sqrt{5}}{3}i,\frac{4}{3}-\frac{\sqrt{5}}{3}i \right\}$.
Work Step by Step
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 $\left\{ \frac{4}{3}+\frac{\sqrt{5}}{3}i,\frac{4}{3}-\frac{\sqrt{5}}{3}i \right\}$.
