Answer
$x=\left\{ \dfrac{2-i}{3},\dfrac{2+i}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
(x-1)(9x-3)=-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}
x(9x)+x(-3)-1(9x)-1(-3)=-2
\\\\
9x^2-3x-9x+3=-2
\\\\
9x^2+(-3x-9x)+(3+2)=0
\\\\
9x^2-12x+5=0
.\end{array}
The quadratic equation above has $a=
9
, b=
-12
, c=
5
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-12)\pm\sqrt{(-12)^2-4(9)(5)}}{2(9)}
\\\\
x=\dfrac{12\pm\sqrt{144-180}}{18}
\\\\
x=\dfrac{12\pm\sqrt{-36}}{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{12\pm\sqrt{-1}\cdot\sqrt{36}}{18}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{12\pm i\sqrt{36}}{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{12\pm i\sqrt{(6)^2}}{18}
\\\\
x=\dfrac{12\pm 6i}{18}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel6(2)\pm \cancel6(1)i}{\cancel6(3)}
\\\\
x=\dfrac{2\pm i}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{2-i}{3}
\\\\\text{OR}\\\\
x=\dfrac{2+i}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{2-i}{3},\dfrac{2+i}{3} \right\}
.$
