Question

How would you find the real solution, by using the quadratic formula.
x^2-4x+2=0

Answer 1

$x^2-4x+2=0 \\ \\ a=1 \\ b=-4 \\ c=2 \\ b^2-4ac=(-4)^2-4 \times 1 \times 2=16-8=8 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4) \pm \sqrt{8}}{2 \times 1}=\frac{4 \pm \sqrt{4 \times 2}}{2}=\frac{4 \pm 2\sqrt{2}}{2}=\frac{2(2 \pm \sqrt{2})}{2}=2 \pm \sqrt{2} \\ \boxed{x=2-\sqrt{2} \hbox{ or } x=2+\sqrt{2}}$

Answer 2

$x^2-4x+2=0\\x^2-4x+4-2=0\\ (x-2)^2=2\\ x-2=\sqrt2 \vee x-2=-\sqrt2\\ x=2+\sqrt2 \vee x=2-\sqrt2$