Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{3-\sqrt{73}}{2},\dfrac{3+\sqrt{73}}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(x-5)(x+2)=6 ,$ 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(2)-5(x)-5(2)=6 \\\\ x^2+2x-5x-10=6 \\\\ x^2-3x-10=6 .\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^2-3x-10-6=0 \\\\ x^2-3x-16=0 .\end{array} The quadratic equation above has $a= 1 , b= -3 , c= -16 .$ 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)(-16)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9+64}}{2} \\\\ x=\dfrac{3\pm\sqrt{73}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{3-\sqrt{73}}{2} \\\\\text{OR}\\\\ x=\dfrac{3+\sqrt{73}}{2} .\end{array} Hence, $x=\left\{ \dfrac{3-\sqrt{73}}{2},\dfrac{3+\sqrt{73}}{2} \right\} .$