Solve the following quatratic equations by completing square method :
`2y^(2)+9y+10=0`

Answer

`-(5)/(2),-2` are the roots of the given quadratic equation.

Explanation

`2y^(2)+9y+10=0`
`:.y^(2)+(9y)/(2)+5=0" "......` (Dividing by 2)
Comparing `y^(2)+(9y)/(2)" with "y^(2)+2yz,`
`2yx=(9y)/(2):.z=(9)/(4)" "z^(2)=(81)/(16)`
`:.y^(2)+(9y)/(2)+(81)/(4)` is a perfect square trinomia.
`y^(2)+(9y)/(2)+5=0`
`:.y^(2)+(9y)/(2)+(81)/(16)-(81)/(16)+5=0`
`:.(y+(9)/(4))^(2)-((81)/(16)-5)=0`
`:.(y+(9)/4)^(2)-((81-80)/(16))=0`
`:.(y+(9)/(4))^(2)-((1)/(16))=0`
`:.(y+(9)/(4))^(2)-((1)/(4))^(2)=0`
`:.(y+(9)/(4)+(1)/(4))(y+(9)/(4)-(1)/(4))=0`
`:.(y+(9+1)/(4))(y+(9-1)/(4))=0`
`:.(y+(10)/(4))(y+(8)/(4))=0`
`:.y+(10)/(4)=0ory+(8)/(4)=0`
`:.y=-(10)/(4)ory=-(8)/(4)`
`:.y=-(5)/(2)ory=-2`