Given the equations:

1. \(y^2 + 2x = 13\)

2. \(2y + x = 8\)

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

\(x = 8 - 2y\)

Substitute \(x = 8 - 2y\) into equation (1):

\(y^2 + 2(8 - 2y) = 13\)

Simplify:

\(y^2 + 16 - 4y = 13\)

\(y^2 - 4y + 3 = 0\)

Factor the quadratic equation:

\((y - 3)(y - 1) = 0\)

Thus, \(y = 3\) or \(y = 1\).

For \(y = 3\):

\(x = 8 - 2(3) = 2\)

Point: \((2, 3)\)

For \(y = 1\):

\(x = 8 - 2(1) = 6\)

Point: \((6, 1)\)

Therefore, the points of intersection are \((2, 3)\) and \((6, 1)\).