## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{-1-\sqrt{29}}{2},\dfrac{-1+\sqrt{29}}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(x+2)(x-3)=1 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} x(x)+x(-3)+2(x)+2(-3)=1 \\\\ x^2-3x+2x-6=1 \\\\ x^2-x-6=1 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2-x-6-1=0 \\\\ x^2-x-7=0 .\end{array} The quadratic equation above has $a= 1 , b= -1 , c= -7 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-1\pm\sqrt{1^2-4(1)(-7)}}{2(1)} \\\\ x=\dfrac{-1\pm\sqrt{1+28}}{2} \\\\ x=\dfrac{-1\pm\sqrt{29}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-1-\sqrt{29}}{2} \\\\\text{OR}\\\\ x=\dfrac{-1+\sqrt{29}}{2} .\end{array} Hence, $x=\left\{ \dfrac{-1-\sqrt{29}}{2},\dfrac{-1+\sqrt{29}}{2} \right\} .$