Question

solve competing the square method.2y^2+9y+10=0​

Answer 1

Answer:

Given : p ( y ) = 2 y² + 9 y + 10 = 0.

⇒ 2 / 2 y² + 9 / 2 y + 10 / 2 = 0 / 2. ⇒ y² + 9 / 2 y + 5 = 0.

Now transfer 5 [ constant term ] R.H.S. side : ⇒ y² + 9 / 2 y = - 5.

i.e. ( 9 / 4 )² = 81 / 16. ...

⇒ ( y + 9 / 4 ) = ± ( 1 / 4 )

Value of y as : ...

y = - 2. ...

y = - 5 / 2