## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $2x^2+3x-1=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a= 2 , b= 3 , c= -1 .$ 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(2)(-1)}}{2(2)} \\\\ x=\dfrac{-3\pm\sqrt{9+8}}{4} \\\\ x=\dfrac{-3\pm\sqrt{17}}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-\sqrt{17}}{4} \\\\\text{OR}\\\\ x=\dfrac{-3+\sqrt{17}}{4} .\end{array} Hence, $x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\} .$