The height of a triangle is 4 in. greater than twice its base. the area of the triangle is no more than 168 in.2. which inequality can be used to find the possible lengths, x, of the base of the triangle?

Recall that A = 1/2bh. We are given that h = 4+2b So, putting it all together: 168 = 1/2 b(4+2b) 168 = 1/2(4b + 2b^2) 168 = 2b + b^2 b^2 + 2b - 168 = 0. Something that multiplies to -168 and adds to 2? There's a trick to this. Notice 13^2 = 169. So, it's more than likely in the middle of the two numbers we're trying to find. So let's try 12 and 14. Yep. 12 x 14 = 168. So this factors into (b+14)(b-12) So b = -14 or b =12. Is it possible to have a negative length on a base? No. So 12 must be our answer. Let's check this. If 12 is our base, then according to our problem, 2*12 + 4 would be our height... or 28. so what is 12 * 28 /2? 196. Check. Hope this helped!

X=base 
(2x+4)=height
Area of a triangle=(base x height)/2
We can suggest this inequality:

x(2x+4)/2<168
x(2x+4)<336
2x²+4x<336
2x²+4x-336<0
x²+2x-168<0

1) we change the inequality sign of inequality by equality sign.
x²+2x-168=0
x=[-2⁺₋√(4+672)] / 2
x=(-2⁺₋√676)/2
we have two solutions:
x₁=(-2-√676)/2=-14
x₂=(-2⁺√676)/2=12

2) we make intervals with these values.
(-∞, -14 )
(-14,12)
(12, +∞)

3) we check  it out if these intervals work
(-∞,-14), For instead; if x=-15⇒ (-15)²+2(-15)-168=27>0. this solution interval don´t work.

(-14,12), For example; if x=1 ⇒(-1)²+2(1)-168=-168<0 this solution interval works.
but the solutions have to be always positive numbers, therefore, the solution is:  (0,12).

(12,+∞) , For example, if x=13 ⇒ (13)²+2(13)-168=27>0, this solution  interval don´t work.

the inequality is : x(2x+4)/2<168, and the interval solution is (0,12) for the values of the base. 

What the heck don’t understand