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

Answer 2

`x^2-4x+2=0 \\ \\ a=1 \\ b=-4 \\ c=2 \\ b^2-4ac=(-4)^2-4 * 1 * 2=16-8=8 \\ \\ x=(-b \pm √(b^2-4ac))/(2a)=(-(-4) \pm √(8))/(2 * 1)=(4 \pm √(4 * 2))/(2)=(4 \pm 2√(2))/(2)=(2(2 \pm √(2)))/(2)=2 \pm √(2) \\ \boxed{x=2-√(2) \hbox{ or } x=2+√(2)}`