Answer
$\left\{-\dfrac{3}{4}\pm\dfrac{\sqrt{23}}{4}i\right\}$
Work Step by Step
The given equation, $
2x^2+3x+4=0
,$ has
\begin{align*}a=
2
,b=
3
,\text{ and }c=
4
.\end{align*}
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-3\pm\sqrt{3^2-4(2)(4)}}{2(2)}
\\\\&=
\dfrac{-3\pm\sqrt{9-32}}{4}
\\\\&=
\dfrac{-3\pm\sqrt{-23}}{4}
.\end{align*}
Using concepts of simplifying radicals, the equation above is equivalent to
\begin{align*}
x&=\dfrac{-3\pm\sqrt{23\cdot(-1)}}{4}
\\\\&=
\dfrac{-3\pm\sqrt{23}\cdot\sqrt{-1}}{4}
\\\\&=
\dfrac{-3\pm\sqrt{23}\cdot i}{4}
&(\text{use }i=\sqrt{-1})
\\\\&=
-\dfrac{3}{4}\pm\dfrac{\sqrt{23}}{4}i
.\end{align*}
Hence, the solution set of $
2x^2+3x+4=0
$ is $
\left\{-\dfrac{3}{4}\pm\dfrac{\sqrt{23}}{4}i\right\}
$.
