Answer
$r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2}
\right\}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
(r-3)(r+5)=2
,$ 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}
r(r)+r(5)-3(r)-3(5)=2
\\\\
r^2+5r-3r-15=2
\\\\
r^2+2r-15=2
.\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}
r^2+2r-15-2=0
\\\\
r^2+2r-17=0
.\end{array}
The quadratic equation above has $a=
1
, b=
2
, c=
-17
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
r=\dfrac{-2\pm\sqrt{2^2-4(1)(-17)}}{2(1)}
\\\\
r=\dfrac{-2\pm\sqrt{4+68}}{2}
\\\\
r=\dfrac{-2\pm\sqrt{72}}{2}
.\end{array}
Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to
\begin{array}{l}\require{cancel}
r=\dfrac{-2\pm\sqrt{36\cdot2}}{2}
\\\\
r=\dfrac{-2\pm\sqrt{(6)^2\cdot2}}{2}
\\\\
r=\dfrac{-2\pm6\sqrt{2}}{2}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
r=\dfrac{\cancel2(-1)\pm\cancel2(3)\sqrt{2}}{\cancel2(1)}
\\\\
r=-1\pm3\sqrt{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
r=-1-3\sqrt{2}
\\\\\text{OR}\\\\
r=-1+3\sqrt{2}
.\end{array}
Hence, $
r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2}
\right\}
.$
