Answer
Solution set: $\{1+\sqrt 5, 1- \sqrt 5\}$
Work Step by Step
$x^{2} = 2x+4$
$x^{2}-2x-4=0$
Using quadratic formula,
$x= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=1, b=-2, c=-4$
$x= \frac{-(-2)±\sqrt ((-2)^{2}-4(1)(-4))}{2(1)}$
$x= \frac{2±\sqrt (4+16)}{2}$
$x= \frac{2±\sqrt 20}{2}$
$x= \frac{2±\sqrt 4\times5}{2}$
$x= \frac{2±2\sqrt 5}{2}$
$x= \frac{2(1±\sqrt 5)}{2}$
$x=1±\sqrt 5$
$x=1+\sqrt 5$ or $x=1-\sqrt 5$
