Another math question...Please tell me what I'm doing wrong...

Express the solution to the following inequality using interval notation.

x2 + 3x - 18 < 0

A. (-6, 3)
B. (-3, 6)
C. (-∞, -6)∪(3, ∞)
D. (-∞, -3)∪(6, ∞)

So here's what I did. I factored the inequality.
(x + 6)(x -3)

I put 6 and -3 on a number line. The points I picked were -4, 0, and 7.

-4^2 + 3(-4) - 18 < 0
-14 < 0 So...True.

0 + 0 -18 < 0
-18 < 0 So...True.

7^2 + 3(7) - 18 < 0
52 < 0 False.

So, the regions on the number line that are correct in interval notation are (-infinity, -3) and (-3, 6), but those together aren't one of the answers. What am I doing wrong?

(x + 6) (x − 3) has roots −6 and 3 (NOT 6 and −3)

So put −6 and 3 on number line, the pick points in each interval:
−7, 0, 4

(−7)^2 + 3(−7) − 18 < 0
49 − 21 − 18 < 0
10 < 0 ... False

0^2 + 3(0) − 18 < 0
−18 < 0 ... True

4^2 + 3(4) − 18 < 0
16 + 12 − 18 < 0
10 < 0 ... False

Solution: (−6, 3)

This IS one of the solutions

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Alternate method: complete the square:

x² + 3x − 18 < 0
x² + 3x < 18
x² + 3x + 9/4 < 18 + 9/4
(x + 3/2)² < 81/4
|x + 3/2| < 9/2
−9/2 < x + 3/2 < 9/2
−9/2 − 3/2 < x < 9/2 − 3/2
−6 < x < 3

Solution using interval notation: (−6, 3)

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ANSWER TO ADDITIONAL DETAILS

You factored correctly. But once you factor, you need to find where expression = 0
Why? Because this is where it will go from > 0 to < 0 (or vice versa)

So we solve (x² + 3x − 18) = 0 ---> (x + 6) (x − 3) = 0, which occurs when
x + 6 = 0 ----> x = −6
or
x − 3 = 0 ----> x = 3

ok factor:
(x + 6)(x - 3) < 0

so tthe possible solutions must be:
-∞ ______________ -6 ___________ 0____3 _______
so lets see: -6 and 3 wont work since it's < and 0 < 0 is false.