1. `3x^2+6x-1`, find Zeros of a polynomial

Solution:
The Given Polynomial `=3x^2+6x-1`

`3x^2+6x-1=0`

`=>3x^2+6x-1 = 0`

factor is not possible for equation `3x^2+6x-1=0`

But we are trying find solution using the method of perfect square.

Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`

we get, `a=3, b=6, c=-1.`

`:. Delta=b^2-4ac`

`=(6)^2-4 (3) (-1)`

`=36+12`

`=48`

`:. sqrt(Delta)=sqrt(48)=4sqrt(3)`



Now, `alpha=(-b+sqrt(Delta))/(2a)`

`=(-(6)+4sqrt(3))/(2*3)`

`=(-6+4sqrt(3))/6`

`=(-3+2sqrt(3))/3`



and, `beta=(-b-sqrt(Delta))/(2a)`

`=(-(6)-4sqrt(3))/(2*3)`

`=(-6-4sqrt(3))/6`

`=(-3-2sqrt(3))/3`


`=>x = (-3+2sqrt(3))/3" or "x = (-3-2sqrt(3))/3`