How do you solve (x-0.2)^2=0.64

Please show me how to solve this, if you can step by step thanks

(x-0.2)^2=0.64
x-0.2=±√0.64
x-0.2=.8
x=1
x-0.2=-0.8
x=-0.6

(x-0.2)^2=0.64
x^2 - 0.4x + 0.04 = 0.64
x^2 - 0.4x - 0.6 = 0
quadratic equation
[-b +/- sqrt(b^2 - 4ac)] / 2a
a = 1, b = -0.4, c = -0.6

x = 1, 0.6

(x - 0.2)^2 = 0.64

(x - 2/10)^2 = 64/100

x - 2/10 = +/- sqrt(64/100)

x - 2/10 = +/- sqrt(64) / sqrt(100)

x - 2/10 = +/- 8/10

x = 2/10 +/- 8/10

x = 2/10 + 8/10 or x = 2/10 - 8/10

x = 10/10 or x = -6/10

x = 1 or x = -3/5

if take both side sqrt

then sqrt ((x-0.2)^2)=sqrt(0.64) so (x-0.2)=0.8 there-four x = 1
believe me about that. im math teacher :)