y=x²-4x-2

It seems like you have a quadratic equation:

\[ y = x^2 - 4x - 2 \]

And you mentioned \( y = x = 2 \). Let's clarify:

1. If you want to find the value of \( y \) when \( x = 2 \), we can substitute \( x = 2 \) into the equation:

\[ y = (2)^2 - 4(2) - 2 \] \[ y = 4 - 8 - 2 \] \[ y = -6 \]

So when \( x = 2 \), \( y = -6 \).

2. If you meant to find the points where \( y = x \), you can set the equation equal to \( x \):

\[ x = x^2 - 4x - 2 \]

Rearranging:

\[ 0 = x^2 - 5x - 2 \]

Now, you can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -5 \), and \( c = -2 \):

\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} \] \[ x = \frac{5 \pm \sqrt{25 + 8}}{2} \] \[ x = \frac{5 \pm \sqrt{33}}{2} \]

This gives you two solutions for \( x \). You can substitute these back into \( y = x \) to find the corresponding \( y \) values.

If you have a different question or requirement, please clarify!