Problem with derivatives

Find the points on the curve y=x^3-x^2-5x+7 where the tangent is parallel to the x-axis?

I would really appreciate if anybody could help me solve this problem.
Thank you!

Tangent lines are parallel to the x-axis when dy/dx = 0.
dy/dx = 3x^2 - 2x - 5
0 = 3x^2 - 2x - 5
0 = (3x - 5)(x + 1)
x = 5/3 or x = -1

y = x³ - x² - 5x + 7
slope of tangent = dy/dx = 3x² - 2x - 5

3x² - 2x - 5 = 0
x = [2 ± √(2² + 4·3·5)]/[2·3]
= [2 ± √64]/6
= [2 ± 8]/6
= -1, 5/3

(-1, 10) and (5/3, 0.519)