Question

What is the solution of the set?

Answer 1

$x^2=16-4x\ \ \ |+4x\\\\x^2+4x=16\ \ \ \ |+2^2\\\\\underbrace{x^2+2\cdot x\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}=16+2^2\\\\(x+2)^2=20\iff x+2=\pm\sqrt{20}\\\\x+2=-\sqrt{4\cdot5}\ \vee\ x+2=\sqrt{4\cdot5}\ \ \ |-2\\\\x=-2-2\sqrt5\ \vee\ x=-2+2\sqrt5$

Answer: D.