Solve 5z^2 is less than or equal to -17z-6?

-3<= z<=(-2/5)

5z^2 ≤ -17z - 6
5z^2 + 17z + 6 ≤ 0
5z^2 + 15z + 2z + 6 ≤ 0
(5z^2 + 15z) + (2z + 6) ≤ 0
5z(z + 3) + 2(z + 3) ≤ 0
(z + 3)(5z + 2) ≤ 0

z + 3 ≥ 0
z ≥ -3

5z + 2 ≤ 0
5z ≤ -2
z ≤ -2/5 (-0.4)

∴ -3 ≤ z ≤ -2/5 (-0.4)

First, make one side of the inequality equal to zero.

5z^2 + 17z + 6 <= 0

Now factor the trinomial.

5z^2 + 15z + 2z + 6 <= 0
5z(z + 3) + 2(z + 3) <= 0
(5z + 2)(z + 3) <= 0

5z + 2 <= 0
z <= 5/2

z + 3 <= 0
z <= -3

5z^2 <= -17z - 6
5z^2 + 17z + 6 <= 0

Now it gets a little tricky. A quadratic can be factored into two factors:
(az + b)(cz + d) <= 0

For a product of two factors to be less than or equal to zero, they have to have opposite signs. One is positive, one is negative. So there are two different cases.

Either:
az + b <= 0 and cz + d >= 0
or
az + b >= 0 and cz + d <= 0

So you have to
1) Factor 5z^2 + 17z + 6
2) Set up these two different cases, each with two inequalities
3) Solve each inequality for z. If both az + b <= 0 and cz + d >= 0 have to be true, are there are any z which satisfy both?
Example: If you got z <= 5 and z >= 7, those can't both be true so you'd say that case has no solutions. If you got z <= 7 and z >= 5, that is true for all z between 5 and 7.

4) Your possible values of z are the ones which satisfy both the first two inequalities (if any) OR both the second two inequalities (if any).