Answer
$x=\left\{ \dfrac{1-i\sqrt{6}}{3},\dfrac{1+i\sqrt{6}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
9x^2-6x=-7
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
9x^2-6x+7=0
.\end{array}
The quadratic equation above has $a=
9
, b=
-6
, c=
7
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(9)(7)}}{2(9)}
\\\\
x=\dfrac{6\pm\sqrt{36-252}}{18}
\\\\
x=\dfrac{6\pm\sqrt{-216}}{18}
.\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}
x=\dfrac{6\pm\sqrt{-1}\cdot\sqrt{216}}{18}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{6\pm i\sqrt{216}}{18}
.\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}
x=\dfrac{6\pm i\sqrt{36\cdot6}}{18}
\\\\
x=\dfrac{6\pm i\sqrt{(6)^2\cdot6}}{18}
\\\\
x=\dfrac{6\pm i(6)\sqrt{6}}{18}
\\\\
x=\dfrac{6\pm 6i\sqrt{6}}{18}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel6(1)\pm \cancel6(1)i\sqrt{6}}{\cancel6(3)}
\\\\
x=\dfrac{1\pm i\sqrt{6}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{1-i\sqrt{6}}{3}
\\\\\text{OR}\\\\
x=\dfrac{1+i\sqrt{6}}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{1-i\sqrt{6}}{3},\dfrac{1+i\sqrt{6}}{3} \right\}
.$
