So you definately wrote something wrong but I will assume the 6x^2 was meant to be just 6x.
2x^2 + 6x + 2 = 0
x^2 + 3x + 1 = 0
I can't seem to find a better way than to simply use the quadratic formula:
(-3 + sqrt(9 - 4*1*1))/2 [tex]\approx[/tex] -0.382
(-3 - sqrt(9 - 4*1*1))/2 [tex]\approx[/tex] -2.62
So the zeroes for this function are at
(-0.382, 0) and (-2.62, 0)
Or exactly:
((-3 + sqrt(9 - 4*1*1))/2, 0) and ((-3 - sqrt(9 - 4*1*1))/2, 0)
