Answer
$\left\{\dfrac{7\pm\sqrt{177}}{4}\right\}$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the FOIL Method, the the given equation, $
(r-5)(2r+3)=1
,$ is equivalent to
\begin{align*}
r(2r)+r(3)-5(2r)-5(3)&=1
\\
2r^2+3r-10r-15&=1
\\
2r^2-7r-15&=1
\\
2r^2-7r+(-15-1)&=0
\\
2r^2-7r-16&=0
.\end{align*}
Using $ax^2+bx+c=0$, the equation above has $a=
2
$, $b=
-7
$, and $c=
-16
$.
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
x&=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(2)(-16)}}{2(2)}
\\\\&=
\dfrac{7\pm\sqrt{49+128}}{4}
\\\\&=
\dfrac{7\pm\sqrt{177}}{4}
.\end{align*}
Hence, the solution set of the equation $
(r-5)(2r+3)=1
$ is $\left\{\dfrac{7\pm\sqrt{177}}{4}\right\}$.
