# Solve by using the quadratic formula. Put answer in exact value, simplest radical form. 3x^2 + 6x + 2 = 0

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`3x^2 + 6x + 2 = 0` This equation is written in `ax^2 + bx + c` form.

Therefore, `a = 3, b = 6, and c = 2`

The quadratic formula is: `(-b+-sqrt(b^2 - 4ac))/(2a)`

Substituting 3, 6, and 2 for a, b, and c respectively we get:

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

Simplyfing in radical we get: `( -6+-sqrt(12)) / 6`

`sqrt(12)` can be simplified: `sqrt(12) =sqrt(4)*sqrt(3)` `rArr` `2sqrt(3)`

Now we have: `( -6+-2sqrt(3)) / 6`

Since all terms have a common factor of 2, we will s**implify and get:**

**`( -3+-sqrt(3) ) / 3` **

The solutions for x in a quadratic equation of the type:

`ax^2+bx+c` is given by

`x=(-b+-sqrt(b^2-4*a*c))/(2*a)`

The given equation is:

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

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

`=(-6+-sqrt(36-24))/(6)`

`=(-6+-2sqrt3)/6`

`=(-3+-sqrt3)/3`

i.e. `x=(-3+sqrt3)/3` , `(-3-sqrt3)/3`