I can't solve this exercice: x^2−(2+√2)x+2√2=0?

x² - (2 + √2)x + 2√2 = 0
x² - 2x - √2x + 2√2 = 0
(x² - 2x) + (-√2x + 2√2) = 0
x(x - 2) - √2(x - 2) = 0
(x - √2)(x - 2) = 0
x = √2  or  x = 2

ax^2 + bx + c = 0
x = (-b +/- sqrt(b^2-4ac))/2a

x^2 - (2+√2)x + 2√2 = 0
x = ((2+√2) +/- √((2+√2)^2-4(2√2)))/2
 = 2, √2

It is just a quadratic equation. Don't let the radicals shake you. Any quadratic equation can be solved with the quadratic formula, but this is a case in which factorization is easier.

x² - [2 + √(2)]x + 2√(2) = 0
[x - √(2)](x - 2) = 0
x = √(2) or x = 2

It's just a quadratic

a = 1
b = -(2 + sqrt(2))
c = 2 * sqrt(2)

b^2 - 4ac =>
(-(2 + sqrt(2)))^2 - 4 * 1 * 2 * sqrt(2) =>
(2 + sqrt(2))^2 - 8 * sqrt(2) =>
4 + 4 * sqrt(2) + 2 - 8 * sqrt(2) =>
4 - 4 * sqrt(2) + 2 =>
(2 - sqrt(2))^2

You can see that relationship with the following identities:

(x + y)^2 = x^2 + 2xy + y^2
(x - y)^2 = x^2 - 2xy + y^2

(-b +/- sqrt(b^2 - 4ac)) / (2a) =>
(2 + sqrt(2) +/- sqrt((2 - sqrt(2))^2)) / (2 * 1) =>
(2 + sqrt(2) +/- (2 - sqrt(2))) / 2

(2 + sqrt(2) + 2 - sqrt(2)) / 2 , (2 + sqrt(2) - 2 + sqrt(2)) / 2
4/2 , 2 * sqrt(2) / 2
2 , sqrt(2)

Test them and watch them work in the original problem.

x^2 - (2 + sqrt(2)) * x + 2 * sqrt(2) = 0

x = 2

2^2 - (2 + sqrt(2)) * 2 + 2 * sqrt(2) =>
4 - 4 - 2 * sqrt(2) + 2 * sqrt(2) =>
0

x = sqrt(2)

2 - (2 + sqrt(2)) * sqrt(2) + 2 * sqrt(2) =>
2 - 2 * sqrt(2) - 2 + 2 * sqrt(2) =>
0

start by factoring...
==> x(x- sqrt(2)) - 2(x - sqrt(2)) = 0
==> (x-2)(x-sqrt(2)) = 0
==> x = 2 or x = sqrt(2)
or use the quadratic formula:

If you look carefully it factorises:

x² − (2 + √2)x + 2√2 = 0

(x - 2)(x - √2) = 0

x  = 2 or x = √2

x=2, x=sqrt(2)

x^2-(2+sqr(2))x+2sqr(2)=0
=>
(x-sqr(2))(x-2)=0
=>
x=sqr(2) or x=2

Cherk:
(x-sqr(2))(x-2)=0
=>
x^2-2x-sqr(2)x+2sqr(2)=0
=>
x^2-(2+sqr(2))x+2sqr(2)=0
valid.

x^2 − (2 + √2)x + 2√2 = 0
-(sqrt(2) - x) (x - 2) = 0
Solutions:
x = 2
x = √2