Factor the trinomial `x^2-11x+24`

`(x-3)(x-8)>=0`

Set each of the factors of the left-hand side of the inequality equal to `0` to find the critical points.

`x-3>=0`

`x-8>=0`

Since `-3` does not contain the variable to solve for, move it to the right-hand side of the inequality by adding `3` to both sides.

`x>=3`

`x-8>=0`

Set each of the factors of the left-hand side of the inequality equal to `0` to find the critical points.

`x>=3`

`x-8>=0`

Since -8 does not contain the variable to solve for, move it to the right-hand side of the inequality by adding 8 to both sides.

`x>=3`

`x>=8`

Since this is a 'greater than `0` ' inequality, all intervals that make the expression positive are part of the solution.

`x<=3or x>=8`

The inequality `x^2-11*x+24 >= 0` has to be solved.

`x^2-11*x+24 >= 0`

First, factor the left hand side

=> `x^2 - 8x - 3x + 24 >= 0`

=> `x(x - 8) - 3(x - 8) >= 0`

=> `(x - 3)(x - 8) >= 0`

This is true in two cases:

- `x - 3 >= 0` and `x - 8 >= 0`

=> `x >= 3` and `x >= 8`

=> `x >= 8`

- `x - 3 <= 3` and `x - 8 <= 8`

=> `x <= 3` and `x <= 8`

=> `x <= 3`

The inequality is met for values of x that lie in `(-oo, 3]U[8, oo)` .

**The solution of `x^2-11*x+24 >= 0` is `(-oo, 3]U[8, oo)` .**