Answer
$r=\left\{ 3-i\sqrt{5},3+i\sqrt{5} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
r^2-6r+14=0
,$ use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
The quadratic equation above has $a=
1
, b=
-6
, c=
14
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
r=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(14)}}{2(1)}
\\\\
r=\dfrac{6\pm\sqrt{36-56}}{2}
\\\\
r=\dfrac{6\pm\sqrt{-20}}{2}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
r=\dfrac{6\pm\sqrt{-1}\cdot\sqrt{20}}{2}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
r=\dfrac{6\pm i\sqrt{20}}{2}
.\end{array}
Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
r=\dfrac{6\pm i\sqrt{4\cdot5}}{2}
\\\\
r=\dfrac{6\pm i\sqrt{(2)^2\cdot5}}{2}
\\\\
r=\dfrac{6\pm i(2)\sqrt{5}}{2}
\\\\
r=\dfrac{6\pm 2i\sqrt{5}}{2}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
r=\dfrac{\cancel2(3)\pm \cancel2(1)i\sqrt{5}}{\cancel2(1)}
\\\\
r=3\pm i\sqrt{5}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
r=3-i\sqrt{5}
\\\\\text{OR}\\\\
r=3+i\sqrt{5}
.\end{array}
Hence, $
r=\left\{ 3-i\sqrt{5},3+i\sqrt{5} \right\}
.$
