The given equations are:

1. \(x^2 - y - 1 = 0\)
2. \(y - 2x + 2 = 0\)

We can solve these equations simultaneously to find the values of \(x\) and \(y\).

From equation (2), we can express \(y\) in terms of \(x\):

\(y = 2x - 2\)

Substituting this into equation (1) gives:

\(x^2 - (2x - 2) - 1 = 0\)
\(x^2 - 2x + 2 - 1 = 0\)
\(x^2 - 2x + 1 = 0\)

This is a quadratic equation in the form \(ax^2 + bx + c = 0\), which has solutions given by the formula:

\(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\)

Substituting \(a = 1\), \(b = -2\), and \(c = 1\) into this formula gives:

\(x = [2 ± sqrt((-2)^2 - 4*1*1)] / (2*1)\)
\(x = [2 ± sqrt(4 - 4)] / 2\)
\(x = [2 ± 0] / 2\)

So, \(x = 1\).

Substituting \(x = 1\) into equation (2) gives:

\(y = 2*1 - 2 = 0\)

So, the solution to the system of equations is (x, y) = (1, 0).