Answer
$x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
x^2-4=2x
,$ express 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}
x^2-2x-4=0
.\end{array}
The quadratic equation above has $a=
1
, b=
-2
, c=
-4
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-4)}}{2(1)}
\\\\
x=\dfrac{2\pm\sqrt{4+16}}{2}
\\\\
x=\dfrac{2\pm\sqrt{20}}{2}
.\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}
x=\dfrac{2\pm\sqrt{4\cdot5}}{2}
\\\\
x=\dfrac{2\pm\sqrt{(2)^2\cdot5}}{2}
\\\\
x=\dfrac{2\pm2\sqrt{5}}{2}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel2(1)\pm\cancel2(1)\sqrt{5}}{\cancel2(1)}
\\\\
x=1\pm\sqrt{5}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=1-\sqrt{5}
\\\\\text{OR}\\\\
x=1+\sqrt{5}
.\end{array}
Hence, $
x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\}
.$
