Answer
$x=\left\{ \dfrac{-1-\sqrt{29}}{2},\dfrac{-1+\sqrt{29}}{2}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$