Algebra: A Combined Approach (4th Edition)

$x_{1}= \sqrt{12} = 2\sqrt{3}$ and $x_{2}= -\sqrt{12} = -2\sqrt{3}$
Given $4x^2-48=0$ $1.)$ Divide by 4 both sides: $4x^2-48=0 \longrightarrow x^2-12=0$ $2.)$ Apply difference of two squares: $x^2-12=0 \longrightarrow (x+\sqrt{12})(x-\sqrt{12})=0$ Therefore the solutions are $x_{1}= \sqrt{12} = 2\sqrt{3}$ and $x_{2}= -\sqrt{12} = -2\sqrt{3}$