What is the solution of x^2+x-6≤0?

1) Factor the polynomial

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

So, can solve (x + 3) (x - 2) ≤ 0

The condition for the product of two factors be zero or negative is that the both are zero or have different signs (one positive and the other negative)

a) when x + 3 ≤ 0 and x - 2 ≥ 0

x + 3 ≤ 0 = > x ≤ - 3

x - 2 ≥ 0 = > x ≥ 2

Given that it is imposible that x be less or equal tha - 3 and greaer or equal than 2, this solution is empy.

b) when x + 3 ≥ 0 and x - 2 ≤ 0

x + 3 ≥ 0 = > x ≥ - 3

x - 2 ≤ 0 = > x ≤ 2

Solution: - 3 ≤ x ≤ 2 or [-3,2]