factorise this: x^2-4x+5=0?

Method 1 :

x² - 4x + 5 = 0
x² - 4x = -5
x² - 4x + 4 = -5 + 4
(x - 2)² = -1
x - 2 = √-1 or x - 2 = -√-1
x - 2 = i or x - 2 = -i
x = 2 + i or x = 2 - i
(where i = √-1)


====
Method 2 :

If x² + bx + c = 0, then x = [-b ± √(b² - 4c)]/2

Now, x² - 4x + 5 = 0
x = {-(-4) ± √[(-4)² - 4(5)]}/2
x = (4 ± √-4)/2
x = (4 ± 2i)/2
x = 2 + i or x = 2 - i
(where i = √-1)

x^2 - 4x + 5 = 0
(x - 2)^2 + 1 = 0
Complex solutions:
x = 2 - i
x = 2 + i

Still does not factorise !!!

x² - 4x + 5 = 0

x² - 4x + 4 + 1 = 0

(x² - 4x + 4) + 1 = 0

(x - 2)² + 1 = 0

(x - 2)² - i² = 0 → you recognize: a² - b² = (a + b).(a - b)

[(x - 2) + i].[(x - 2) - i] = 0

(x - 2 + i).(x - 2 - i) = 0

Jesus is always the answer





No but anyway get an app such as Photomath it can answer stuff for you