## Algebra 2 Common Core

$x=\left\{ \dfrac{-3-\sqrt{61}}{2},\dfrac{-3+\sqrt{61}}{2} \right\}$
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 $x=\left\{ \dfrac{-3-\sqrt{61}}{2},\dfrac{-3+\sqrt{61}}{2} \right\} .$