Help with finding points on curve y=x^3-8x with the gradient of the curve being 4

Question:
Find the coordinates of the points on the curve with the equation y=x^3-8x at which the gradient of the curve is 4.
Could you please walk me through it?

y = x^3 - 8x <=== curve

y ' = 3x^2 - 8 <=== slope of curve

3x^2 - 8 = 4 . . . given slope = 4 ... solve for x

3x^2 - 12 = 0

x^2 - 4 = 0

(x + 2) (x - 2) = 0

x = 2 ; and ; x = - 2

y = 2^3 - 8*2 = - 8 ... (2 , - 8) <===== point on curve with slope of 4
and
y = (-2)^3 - 8*(-2) = 8 ... (- 2 , 8) <===== point on curve with slope of 4

if you have an equation in the form y = some function of x, then if you differentiate you will find the gradient.

the differential rule is:

y = x^n

dy/dx = nx^(n-1)

so y = x^3 - 8x

dy/dx = 3x^2 - 8x^0 = 3x^2 - 8

now if we know that the gradient is 4 we can say that dy/dx = 4

therefore

4 = 3x^2 - 8

3x^2 = 12

divide both sides by three to get

x^2 = 4

therefore x = 2 or x = -2

put these in for y:

x = 2

y = 2^3 - 8(2)

y = -8

so out first coordinate is (2,-8)

x = -2

y = -2^3 -8(-2)

y = 8

this coordinate is (-2,8)

so our points are (2,-8) and (-2,8)