Answer
$\left\{\dfrac{-1-\sqrt{13}}{2}, \dfrac{-1+\sqrt{13}}{2}\right\}$
Work Step by Step
Using the standard form $ax^2+bx+c=0$ as reference, the given quadratic equation has $a=1, b=1, \text{ and } c=-3$.
Solve using the Quadratic Formula $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ using the values above to obtain:
\begin{align*}
x&=\frac{-1\pm\sqrt{1^2-4(1)(-3)}}{2(1)}\\\\
&=\frac{-1\pm\sqrt{1+12}}{2}\\\\
&=\frac{-1\pm\sqrt{13}}{2}\\\\
\end{align*}
Thus, the solution set is:
$\left\{\dfrac{-1-\sqrt{13}}{2}, \dfrac{-1+\sqrt{13}}{2}\right\}$
