Please help me solve 2x + 1 < 8x^2 (squared)

I have no idea about polynomial functions. This is due tomorrow and I really need this broken down. Can anyone help me with this and explain it?

Move everything to one side:

8x^2 - 2x - 1 > 0

Factor it out:

(4x + 1)(2x - 1) > 0

The only way that two numbers can be multiplied together and be greater than zero is if they either both positive or both negative, we have to account for both:

Positive:

4x + 1 > 0
4x > -1
x > -1/4

2x - 1 > 0
2x > 1
x > 1/2

Negative:

4x + 1 < 0
4x < -1
x < -1/4

2x - 1 < 0
2x < 1
x < 1/2

Now we start simplifying our inequalities, we always choose the stronger inequality (the inequality with fewer options). In the positive it says that x is greater than 1/2 and also greater than -1/4, if it is greater than 1/2 then it is certainly greater than -1/4 so use that inequality instead. In the negative it says that x is less than -1/4 and also less than 1/2, if x is less than -1/4 then it is definitely less than 1/2 so take that inequality.

x > 1/2 OR x < -1/4

Given that : 2x + 1 < 8x^2,

ADD -2x-1, both-sides, 0 < 8x^2 -2x -1,
OR,
0 < (2x-1)(4x+1)
SO,
0 < 2x - 1, OR, 0 < 4x+1,

1 < 2x, OR, -1 < 4x,

1/2 < x, OR, -1/4 < x
Thus,
x > 1/2 >==============================< ANSWER