Answer
$x=\left\{
\dfrac{-3-\sqrt{17}}{2},\dfrac{-3+\sqrt{17}}{2}
\right\}
$
Work Step by Step
Using $ax^2+bx+c=0,$ the given equation,
\begin{align*}
x^2+3x-2=0
,\end{align*}
the values of $a,b, \text{ and }c $ are $a=
1
,$ $b=
3
,$ and $c=
-2
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}
x&=\dfrac{-3\pm\sqrt{3^2-4(1)(-2)}}{2(1)}
\\\\&=
\dfrac{-3\pm\sqrt{9+8}}{2}
\\\\&=
\dfrac{-3\pm\sqrt{17}}{2}
.\end{align*}
The solutions are $
x=\left\{
\dfrac{-3-\sqrt{17}}{2},\dfrac{-3+\sqrt{17}}{2}
\right\}
.$
