Question

I don't know what method I should use to solve for y in this equation: (y-3)^2=4y-12

Answer 1

(y-3)^2 -4y +12 = 0 
[y^2 -6y +9] -4y +12 =0 
y^2 -10y + 21 = 0 
(y -3) (y-7) =0 
y=3 ; y = 7