Answer
$\{2-\sqrt{2},2+\sqrt{2}\}$
Work Step by Step
The equation is in standard form $ax^2+bx+c=0$.
We have $a=1,b=-4$ and $c=2$.
The discriminant is
$b^2-4ac$
$=(-4)^2-4(1)(2)$
$=16-8$
$=8$
Since $8>0$. There are two real solutions.
The quadratic formula is
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substitute the values of $a, b, c,$ and the discriminant into the quadratic formula to obtain:
$x=\dfrac{-(-4)\pm \sqrt{(8)}}{2(1)}$
$x=\dfrac{4\pm \sqrt{4(2)}}{2}$
Simplify.
$x=\dfrac{4\pm 2\sqrt{2}}{2}$
Factor out $2$.
$x=\dfrac{2(2\pm \sqrt{2})}{2}$
Simplify.
$x=2\pm \sqrt{2}$
The solution set is $\{2-\sqrt{2},2+\sqrt{2}\}$.
