## Precalculus (6th Edition)

$\color{blue}{\left\{\dfrac{3-\sqrt{17}}{2}, \dfrac{3+\sqrt{17}}{2}\right\}}$.
RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-3, c=-2$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-2)}}{2(1)} \\x=\dfrac{3\pm\sqrt{9+8}}{2} \\x=\dfrac{3\pm\sqrt{17}}{2}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{3-\sqrt{17}}{2}, \dfrac{3+\sqrt{17}}{2}\right\}}$.