Equation of a tangent to a curve calculus help!

So for the following question, I have gone through the steps to solve it. I am just not confident in my answer so if anyone could help that would be appreciated!

Q. Find the equation of the tangent to y = x^2 - 3x - 4 that is parallel to the line y = 7x + 3.

A.

1. The slope of y=7x+3 is 7, hence the slope of the tangent is 7.

2. Finding the point on the curve where the derivative is 7.

y'= 2x - 3

7= 2x -3

7-3 =2x

4=2x

x =2

3. Solving for y;

y= (2)^2 - 3(2) - 4

y = -6

4. The point on the curve is (2, -6)

5. Using the coordinates to find the y-intercept:

y= 7x + b

-6 = 7(2) + b

b= -20

6. The equation of the tangent is y = 7x - 20

Am I way off, or at least on the right track??? Thank you!