Translate the following statements into algebraic equations and solve for x and y: x is the sum of 2 and y. Five times the product of x and y, is 6 more than the square of x.

The first statement says "$x$ is the sum of 2 and $y$". This can be written as:

$$x = 2 + y \quad (1)$$

The second statement says "Five times the product of $x$ and $y$, is 6 more than the square of $x$". This can be written as:

$$5xy = x^2 + 6 \quad (2)$$

From equation (1), we can express $y$ in terms of $x$:

$$y = x - 2$$

Now, substitute this expression for $y$ into equation (2):

$$5x(x - 2) = x^2 + 6$$

Expand the left side of the equation:

$$5x^2 - 10x = x^2 + 6$$

Rearrange the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

$$5x^2 - x^2 - 10x - 6 = 0$$ $$4x^2 - 10x - 6 = 0$$

Divide the entire equation by 2 to simplify:

$$2x^2 - 5x - 3 = 0$$

Factor the quadratic equation:

$$(2x + 1)(x - 3) = 0$$

This gives two possible values for $x$:

$$2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ $$x - 3 = 0 \implies x = 3$$

Using the relation $y = x - 2$:

Case 1: For $x = 3$

$$y = 3 - 2$$ $$y = 1$$

Case 2: For $x = -\frac{1}{2}$

$$y = -\frac{1}{2} - 2$$ $$y = -\frac{1}{2} - \frac{4}{2}$$ $$y = -\frac{5}{2}$$

The values of $x$ and $y$ are:
For the first solution: $\boxed{\text{x = 3, y = 1}}$
For the second solution: \boxed{\text{x = -\frac{1}{2}, y = -\frac{5}{2}}}