Question

Solve x2-14x+21=0 by completing the square

Answer 1

$x^2-14x+21=0\\\\x^2-2x\cdot7+7^2-7^2+21=0\\\\(x-7)^2-49+21=0\\\\(x-7)^2-28=0\\\\(x-7)^2=28\iff x-7=-\sqrt{28}\ \vee\ x-7=\sqrt{28}\\\\x=7-\sqrt{4\cdot7}\ \vee\ x=7+\sqrt{4\cdot7}\\\\x=7-2\sqrt7\ \vee\ x=7+2\sqrt7$