What is the total number of pairs of integers (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4?

First, you'll have to provide an equation. You've provided
an expression, for the moment. We'll transform the expression into an
equation:

x^2-4xy+5y^2+2y-4 =
0

We'll create the following
groups;

(x^2-4xy+4y^2) + (y^2+2y + 1)- 1 - 4 =
0

We notice that we've made some changes within the
expression you've provided, namely:

5y^2 = 4y^2 +
y^2

- we've added and subtracted the value of
1

We notice that creating the groups above, we've created,
in fact, two perfect squares:

(x-2y)^2 + (y+1)^2 - 5 =
0

Now, we'll have to determine what are the integers which
added to give 5?

1 + 4, 2 + 3, 3 + 2, 4 +
1

Let x - 2y = 1 and y + 1 =
2

x = 2y + 1

y = 1 => x
= 3

The first integer pair which satisfies the given
equation is (3,1).

Let x - 2y = sqrt2 and y + 1 =
sqrt3

Since x and y are not integer, we'll move to the next
possibility:

x - 2y = 2 and y + 1 = 1 => y =
0

x = 2

The next possible pair
of integer numbers, that makes the equality to hold is
(2;0).

Therefore, the pairs of integer
numbers that satisfy the given equation are: (3,1) and
(2;0).