Answer
$x=\left\{
\dfrac{-3-\sqrt{61}}{2},\dfrac{-3+\sqrt{61}}{2}
\right\}
$
Work Step by Step
In the given equation,
\begin{align*}
x^2+3x-13=0
,\end{align*} $a=
1
,$ $b=
3
,$ and $c=
-13
.$
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)(-13)}}{2(1)}
\\\\&=
\dfrac{-3\pm\sqrt{9+52}}{2}
\\\\&=
\dfrac{-3\pm\sqrt{61}}{2}
.\end{align*}
Hence, the solutions are $
x=\left\{
\dfrac{-3-\sqrt{61}}{2},\dfrac{-3+\sqrt{61}}{2}
\right\}
.$
