Question

Find roots of 2y²-3y+1=0 by completing square method

Answer 1

Answer:

(2y-1)(y-1) this is correct slove

Answer 2

Step-by-step explanation:

let x=2y^2-3y+1=0

=y^2-3y/2+1/2

=y^2-3/2y+(6/2y)^2-(6/2y)^2+1/2

=(y-3)^2-9+1/2

=(y-3)^2+(-18+1/2)

=(y-3)^2-17/2