Answer
$r=\left\{ \dfrac{13-\sqrt{163}}{3},\dfrac{13+\sqrt{163}}{3}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
26r-2=3r^2
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-3r^2+26r-2=0
\\\\
\dfrac{-3r^2+26r-2}{-1}=\dfrac{0}{-1}
\\\\
3r^2-26r+2=0
.\end{array}
The quadratic equation above has $a=
3
, b=
-26
, c=
2
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
r=\dfrac{-(-26)\pm\sqrt{(-26)^2-4(3)(2)}}{2(3)}
\\\\
r=\dfrac{26\pm\sqrt{676-24}}{6}
\\\\
r=\dfrac{26\pm\sqrt{652}}{6}
.\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{26\pm\sqrt{4\cdot163}}{6}
\\\\
r=\dfrac{26\pm\sqrt{(2)^2\cdot163}}{6}
\\\\
r=\dfrac{26\pm2\sqrt{163}}{6}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
r=\dfrac{\cancel2(13)\pm\cancel2(1)\sqrt{163}}{\cancel2(3)}
\\\\
r=\dfrac{13\pm\sqrt{163}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
r=\dfrac{13-\sqrt{163}}{3}
\\\\\text{OR}\\\\
r=\dfrac{13+\sqrt{163}}{3}
.\end{array}
Hence, $
r=\left\{ \dfrac{13-\sqrt{163}}{3},\dfrac{13+\sqrt{163}}{3}
\right\}
.$
