Answer
$x=7$ or $x=3$
Work Step by Step
Isolate the radical on the right side to obtain:
$x=2\sqrt{x-3} + 3
\\x-3 = 2\sqrt{x-3}$
Square both sides of the equation to obtain:
$(x-3)^2=2^2(x-3)$
Use the rule $(a-b)^2 = a^2 - 2ab+b^2$ where $a=x$ and $b=3$ to obtain:
$\\x^2-6x+9=4(x-3)
\\x^2-6x+9=4x-12$
Move all terms to the left side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite.
$x^2-6x+9-4x+12=0
\\x^2-10x+21=0$
Factor the trinomial to obtain:
$(x-7)(x-3)=0$
Equate each factor to zero then solve each equation to obtain:
$x-7 = 0 \text{ or } x -3 = 0
\\x = 7 \text{ or } x = 3$
