What are the zero coordinates of this equasion?

2x^2 - 6  = 0

2x^2 - (0)x -6 =0 is a quadratic equation   ...use quadratic formula to find 

x= +-\(\sqrt{3}\)

Quadratic Formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

I assume by "zero coordinates", you mean the coordinates of the x-intercepts or "zeroes" of the equation.

The first thing that can be done to find the zeroes is to factor out a 2 from both terms:

2x^2 - 3 = 0

2(x^2 - 3) = 0

x^2 - 3 = 0/2

x^2 - 3 = 0

Now, there are a couple of ways to go from here, but the simplest is probably adding 3 to both sides:

x^2 - 3 = 0

x^2 + 3 - 3 = 0 + 3

x^2 + (3 - 3) = 3

x^2 + 0 = 3

x^2 = 3

Then we can find the square root of both sides to isolate x:

x^2 = 3

√(x^2) = ±√3

It's important to remember that there's a plus or minus before the non-x term.

√(x^2) = ±√3

x = ±√3

x = √3 and x = -√3

Those are the roots.

It's simpler not to bring in the quadratic formula in this case, since there's no "b" term.

It's probably better to understand the simplest way first before trying to use the formula to solve.

Thank you everone!