## College Algebra (11th Edition)

$x=\left\{ \dfrac{3-\sqrt{17}}{2},\dfrac{3+\sqrt{17}}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $x^2-3x-2=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ In the equation above, $a= 1 ,$ $b= -3 ,$ and $c= -2 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(-2)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9+8}}{2} \\\\ x=\dfrac{3\pm\sqrt{17}}{2} .\end{array} The solutions are $x=\left\{ \dfrac{3-\sqrt{17}}{2},\dfrac{3+\sqrt{17}}{2} \right\} .$