Answer
$\{1-\sqrt 5,1+\sqrt 5\}$
Work Step by Step
We have to solve the equation:
$$x^2=2x+4.$$
First we bring the equation to the standard form by moving all terms on one side:
$$x^2-2x-4=0.$$
To solve the equation $ax^2+bx+c=0$ we will use the quadratic formula:
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Identify $a$, $b$, $c$:
$$\begin{align*}
a&=1\\
b&=-2\\
c&=-4.
\end{align*}$$
We solve the given equation by substituting the values of $a$, $b$, $c$ in the quadratic formula:
$$\begin{align*}
x&=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-4)}}{2(1)}\\
&=\dfrac{2\pm\sqrt{20}}{2}\\
&=\dfrac{2\pm 2\sqrt 5}{2}\\
&=1\pm\sqrt 5.
\end{align*}$$
The solution set of the equation is:
$$\{1-\sqrt 5,1+\sqrt 5\}.$$
