solve `x^(2)+6x+6=0`

Explanation

The given equation is `x^(2)+6x+6=0.`
Comparing it with `ax^(2)+bx+c=0,` we get
`a-1,b=6" and "c=6.`
`:." "D=(b^(2)-4ac)=(36-4xx1xx6)=12gt0.`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt(12)=2sqrt(3).`
`:." "alpha=(-b+sqrt(D))/(2a)=((-6+2sqrt(3)))/(2xx1)=((-6+2sqrt(3)))/(2)=(-3+sqrt(3)),`
`beta=(-b-sqrt(D))/(2a)=((-6-2sqrt(3)))/(2xx1)=((-6-2sqrt(3)))/(2)=(-3-sqrt(3)).`
Hence, `(-3+sqrt(3))` and `(-3-sqrt(3))` are the roots of the given equation.