Question 37186
1. Visually, we are looking at a parabola, so it looks like a U with the x-axis going through it horizontally, and hitting it at 2 points.  We want to find those 2 points (the values for x where y=0).
This is a quadratic equation, so we can solve it using the quadratic formula.
x=(-b+/-sqrt(b^2-4ac))/2a
a=3, b=-2, c=-2
x=(-2+/-sqrt(4+24))/6 = (-2+/-sqrt28)/6 = (-2+/-2sqrt7)/6 = (-1+/-sqrt7)/3.

2. For this one, we can rewrite it as 2y^2-y-6=0.
The fact that we have y instead of x makes absolutely no difference.
You can solve it using the same formula as we used above:
a=2, b=-1, c=-6, so y=(1+/-sqrt(1+48))/4 = (1+/-7)/4 = 8/4, -6/4,
So y=2 and y=-3/2.