Solve the following quadratic equations by using formula method :
`y^(2)+(1)/(3)y=2`

Answer

`x=(-1pmsqrt73)/(6),(-1-sqrt73)/(6)` are the roots of the given quadratic equation.

Explanation

`y^(2)+(1)/(3)y=2`
`:.y^(2)+(1)/(3)y-2=0`
Comparing with the standard form `ax^(2)+bx+c=0`
`a=1,b=(1)/(3),c=-2`,
`b^(2)-4ac=((1)/(3))^(2)-4(1)(-2)=(1)/(9)+8`
`=(1+72)/(9)=(73)/(9)`
`x=(-bpmsqrt(b^(2)-4ac))/(2a)=((-1)/(3)pmsqrt((73)/(9)))/(2xx1)`
`=(-(1)/(3)pm(1)/(3)sqrt73)/(2)`
`=(-1pmsqrt73)/(6)`
`:.x=(-1pmsqrt73)/(6)orx=(-1-sqrt73)/(6)`