The given curve is represented by the equation Y = x^2 - x. We need to find the slope of the tangent to this curve at the point where the line Y = 6 intersects the curve in the first quadrant.

To find the point of intersection, we'll set the equation of the curve equal to the equation of the line:
6 = x^2 - x

Now, let's solve this equation to find the x-coordinate of the point of intersection.

Step 1: Rearrange the equation:
x^2 - x - 6 = 0

Step 2: Factorize the quadratic equation:
(x - 3)(x + 2) = 0

Step 3: Set each factor equal to zero and solve for x:
x - 3 = 0 or x + 2 = 0
x = 3 or x = -2

Since we are interested in the first quadrant, we'll consider the positive value of x, which is x = 3.

So, the point of intersection is (3, 6).

Now, let's find the slope of the tangent at this point.

Step 1: Differentiate the equation of the curve with respect to x to find the derivative:
d/dx (Y) = d/dx (x^2 - x)
Y' = 2x - 1

Step 2: Substitute the x-coordinate of the point of intersection into the derivative to find the slope at that point:
Slope = Y'(3) = 2(3) - 1 = 6 - 1 = 5

Therefore, the slope of the tangent to the curve at the point where the line Y = 6 intersects the curve in the first quadrant is 5.

In conclusion, the slope of the tangent is 5.