Answer
$x=\left\{ \dfrac{1-i\sqrt{6}}{2},\dfrac{1+i\sqrt{6}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
4x^2-4x=-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}
4x^2-4x+7=0
.\end{array}
The quadratic equation above has $a=
4
, b=
-4
, c=
7
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(7)}}{2(4)}
\\\\
x=\dfrac{4\pm\sqrt{16-112}}{8}
\\\\
x=\dfrac{4\pm\sqrt{-96}}{8}
.\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{4\pm\sqrt{-1}\cdot\sqrt{96}}{8}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{4\pm i\sqrt{96}}{8}
.\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{4\pm i\sqrt{16\cdot6}}{8}
\\\\
x=\dfrac{4\pm i\sqrt{(4)^2\cdot6}}{8}
\\\\
x=\dfrac{4\pm i(4)\sqrt{6}}{8}
\\\\
x=\dfrac{4\pm 4i\sqrt{6}}{8}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel4(1)\pm \cancel4(1)i\sqrt{6}}{\cancel4(2)}
\\\\
x=\dfrac{1\pm i\sqrt{6}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{1-i\sqrt{6}}{2}
\\\\\text{OR}\\\\
x=\dfrac{1+i\sqrt{6}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{1-i\sqrt{6}}{2},\dfrac{1+i\sqrt{6}}{2} \right\}
.$
