Answer
The solution of the equation $2{{x}^{2}}+2x+3=0$ in standard form is $\left\{ -\frac{1}{2}+\frac{\sqrt{5}}{2}i,-\frac{1}{2}-\frac{\sqrt{5}}{2}i \right\}$.
Work Step by Step
Compare the equation $2{{x}^{2}}+2x+3=0$ with $a{{x}^{2}}+bx+c$.
$\begin{align}
& a=2 \\
& b=2 \\
& c=3
\end{align}$
Substitute $a=2$, $b=2$ and $c=3$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
& x=\frac{-2\pm \sqrt{{{2}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\
& =\frac{-2\pm \sqrt{4-24}}{4} \\
& =\frac{-2\pm \sqrt{-20}}{4}
\end{align}$
Use the property $\sqrt{-b}=i\sqrt{b}$.
\[\begin{align}
& x=\frac{-2\pm i\sqrt{20}}{4} \\
& =\frac{-2\pm i\sqrt{4\cdot 5}}{4} \\
& =\frac{-2\pm 2i\sqrt{5}}{4}
\end{align}\]
Express the complex number in the standard form.
\[\begin{align}
& x=-\frac{2}{4}\pm \frac{2\sqrt{5}}{4}i \\
& =-\frac{1}{2}\pm \frac{\sqrt{5}}{2}i
\end{align}\]
Therefore, the solution of the equation $2{{x}^{2}}+2x+3=0$ in standard form is $\left\{ -\frac{1}{2}+\frac{\sqrt{5}}{2}i,-\frac{1}{2}-\frac{\sqrt{5}}{2}i \right\}$.
