(x-3)^2 >=(x+6)*(x-1)

To solve the inequality (x - 3)^2 ≥ (x + 6)(x - 1), we'll start by expanding both sides of the inequality and then bringing all terms to one side to form a quadratic inequality.

Expanding the left side of the inequality:

(x - 3)(x - 3) = x^2 - 3x - 3x + 9 = x^2 - 6x + 9

Expanding the right side of the inequality:

(x + 6)(x - 1) = x^2 - x + 6x - 6 = x^2 + 5x - 6

Now we have:

x^2 - 6x + 9 ≥ x^2 + 5x - 6

Subtract x^2 from both sides:

-6x + 9 ≥ 5x - 6

Now, we want to move all the terms to one side of the inequality:

-6x - 5x ≥ -6 - 9

-11x ≥ -15

Now, we'll divide both sides by -11. Remember that when we divide by a negative number, we have to flip the inequality sign:

x ≤ -15 / -11

x ≤ 15/11

So the solution to the inequality is:

x ≤ 15/11

This means that all values of x that are less than or equal to 15/11 satisfy the original inequality.