## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{3-i\sqrt{15}}{2},\dfrac{3+i\sqrt{15}}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x^2-3x+6=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 1 , b= -3 , c= 6 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(6)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9-24}}{2} \\\\ x=\dfrac{3\pm\sqrt{-15}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{3\pm\sqrt{-1}\cdot\sqrt{15}}{2} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{3\pm i\sqrt{15}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{3-i\sqrt{15}}{2} \\\\\text{OR}\\\\ x=\dfrac{3+i\sqrt{15}}{2} .\end{array} Hence, $x=\left\{ \dfrac{3-i\sqrt{15}}{2},\dfrac{3+i\sqrt{15}}{2} \right\} .$