We begin by solving the equation for its zeros. This is done by changing the \(\displaystyle \geq\) sign into an \(\displaystyle =\) sign. 

\(\displaystyle x^2+3x=4\Rightarrow x^2+3x-4=0\)

\(\displaystyle \\ \Rightarrow(x-1)(x+4)=0\\ \Rightarrow x=-4,1\)

Since we know the zeros of the equation, we can then check the areas around the zeros since we naturally have split up the real line into three sections :

\(\displaystyle (-\infty,-4),(-4,1),(1,\infty)\)

First we check \(\displaystyle x=-5\)

\(\displaystyle (-5)^2+3(-5)-4=25-15-4=6\geq0\)

Therefore, the first interval can be included in our answer. Additionally, we know that \(\displaystyle x=-4\) satisfies the equation, therefore we can say with certainty that the interval \(\displaystyle (-\infty,-4]\) is part of the answer. 

Next we check something in the second interval. Let \(\displaystyle x=0\), then

\(\displaystyle 0^2+3(0)-4=-4< 0\)

Therefore the second interval cannot be included in the answer.

Lastly, we check the third interval. Let \(\displaystyle x=2\), then

\(\displaystyle 2^2+3(2)-4=4+6-4=6\ge 0\)

Which does satisfy the original equation. Therefore the third interval can also be included in the answer. Since we know that \(\displaystyle x=1\) satisfies the equation as well, we can include it in the interval as such: \(\displaystyle [1,\infty)\)

Therefore, 

\(\displaystyle x=(-\infty,-4]\cup[1,\infty)\)

Method 1:

1) Multiply-out the left side then rewrite the inequality as an equation:

    \(\displaystyle x^{2} + x = 12\)

2) Now rewrite as a quadratic equation and solve the equation:

    \(\displaystyle x^{2} + x - 12 = 0\)

    \(\displaystyle (x + 4)(x - 3) = 0\)

    \(\displaystyle x + 4 = 0\)     \(\displaystyle x - 3 = 0\)

    \(\displaystyle x = -4\)       \(\displaystyle x = 3\)

3) Next set up intervals using the solutions and test the original inequality to   see where it holds true by using values for \(\displaystyle x\) on each interval.

4) The interval between \(\displaystyle x = -4\) and \(\displaystyle x = 3\) holds true for the original inequality.

5) Solution: \(\displaystyle -4 < x < 3\)

Method 2:

Using a graphing calculator, find the graph.  The function is below the x-axis (less than \(\displaystyle 0\)) for the x-values \(\displaystyle -4 < x < 3\).  Using interval notation for \(\displaystyle x\), \(\displaystyle (-4, 3)\).

Method 3:

For the inequality \(\displaystyle x(x + 1) < 12\), the variable expression in terms of \(\displaystyle x\) is less than \(\displaystyle 12\), and an inequality has a range of values that the solution is composed of.  This means that each of the solution values for \(\displaystyle x\) are strictly between the two solutions of \(\displaystyle x(x + 1) = 12\).  'Between' is for a 'less than' case, 'Outside of' is for a 'greater than' case.

1) Write \(\displaystyle -6 \le 6x + 3 \le 21\) as two simple inequalities:

   \(\displaystyle -6 \leq 6x + 3\)          \(\displaystyle 6x + 3 \leq 21\)

2) Solve the inequalities:

   \(\displaystyle -3 - 6 \leq 6x\)            \(\displaystyle 6x \leq 21 - 3\)

   \(\displaystyle -9 \leq 6x\)                   \(\displaystyle 6x \leq 18\)

   \(\displaystyle -\frac{9}{6} \leq x\)                    \(\displaystyle x \leq 3\)

3) Write the final solution as a single compound inequality:

   \(\displaystyle -\frac{9}{6} \leq x \leq 3\)

 For interval notation:

 \(\displaystyle [-\frac{9}{6}, 3] = [-1 \frac{3}{6}, 3]\)

4) Now graph:

In order to solve this equation, we must first isolate the absolute value. In this case, we do it by dividing both sides by \(\displaystyle 3\) which leaves us with:

\(\displaystyle \left | x-6\right |< 7\)\(\displaystyle \left | x-6 \right|< 7\)

When we work with absolute value equations, we're actually solving two equations. So, our next step is to set up these two equations: 

\(\displaystyle x-6< 7\) and \(\displaystyle x-6>-7\)

In both cases we solve for \(\displaystyle x\) by adding \(\displaystyle 6\) to both sides, leaving us with

\(\displaystyle x< 13\) and \(\displaystyle x>-1\)

This can be rewritten as \(\displaystyle -1< x< 13\)

When we work with absolute value equations, we're actually solving two equations: 

\(\displaystyle 4x-2< 10\) and \(\displaystyle 4x-2>-10\)

Adding \(\displaystyle 2\) to both sides leaves us with: 

\(\displaystyle 4x< 12\) and \(\displaystyle 4x>-8\)

Dividing by \(\displaystyle 4\) in order to solve for \(\displaystyle x\) allows us to reach our solution:

\(\displaystyle x< 3\) and \(\displaystyle x>-2\)

Which can be rewritten as:

\(\displaystyle -2< x< 3\)

In order to solve for \(\displaystyle x\) we must first isolate the absolute value. In this case, we do it by dividing both sides by 2:

\(\displaystyle \left | 3x-5\right |>23\)\(\displaystyle |3x-5|>23\)

As with every absolute value problem, we set up our two equations:

\(\displaystyle 3x-5>23\) and \(\displaystyle 3x-5< -23\)

We isolate \(\displaystyle x\) by adding \(\displaystyle 5\) to both sides:

\(\displaystyle 3x>28\) and \(\displaystyle 3x< -18\)

Finally, we divide by \(\displaystyle 3\):

\(\displaystyle x>\frac{28}{3}\) and \(\displaystyle x< -6\)

Our first step in solving this equation is to isolate the absolute value. We do this by dividing both sides by \(\displaystyle 5:\)

\(\displaystyle |\frac{x}{2}+4|< 2\).

We then set up our two equations:

\(\displaystyle \frac{x}{2}+4< 2\)\(\displaystyle \frac{x}{2}+4< 2\) and \(\displaystyle \frac{x}{2}+4>-2\)\(\displaystyle \frac{x}{2}+4>-2\).

Subtracting 4 from both sides leaves us with

\(\displaystyle \frac{x}/{2} < -2\)\(\displaystyle \frac{x}{2}< -2\) and \(\displaystyle \frac{x}{2}>-6\)\(\displaystyle \frac{x}{2} >-6\).

Lastly, we multiply both sides by 2, leaving us with \(\displaystyle x\):

\(\displaystyle x< -4\) and \(\displaystyle x>-12\).

Which can be rewritten as:

\(\displaystyle -12< x < -4\)

We first need to isolate the absolute value, which we can do in two steps:

1. Add 2 to both sides:

\(\displaystyle 4\left | 5x+8\right |< 24\)

2. Divide both sides by 4:

\(\displaystyle \left | 5x+8\right |< 6\)\(\displaystyle |5x+8|< 6\)

Our next step is to set up our two equations:

\(\displaystyle 5x+8< 6\) and \(\displaystyle 5x+8>-6\)

We can now solve the equations for \(\displaystyle x\) by subtracting both sides by 8:

\(\displaystyle 5x< -2\) and \(\displaystyle 5x>-14\)

and then dividing them by 5:

\(\displaystyle x< -\frac{2}{5}\) and \(\displaystyle x>-\frac{14}{5}\)

Which can be rewritten as: 

\(\displaystyle -\frac{14}{5}< x< -\frac{2}{5}\)

\(\displaystyle 7+\left | 4x+13\right |< 36\)

First we need to get the expression with the absolute value sign by itself on one side of the inequality. We can do this by subtracting seven from both sides.

\(\displaystyle \left | 4x+13\right |< 29\)

Next we need to set up two inequalities since the absolute value sign will make both a negative value and a positive value positive.

\(\displaystyle -29< 4x+13< 29\)

From here, subtract thirteen from both sides and then divide everything by four.

\(\displaystyle -29-13< 4x< 29-13\)

\(\displaystyle -42< 4x< 16\)

\(\displaystyle -10.5< x< 4\)

\(\displaystyle 3\left | \frac{2x}{5} +7\right |>18\)

First we need to get the expression with the absolute value sign by itself on one side of the inequality. We can do this by dividing both sides by three. 

\(\displaystyle \left | \frac{2x}{5} +7\right |>6\)

We now have two equations:

\(\displaystyle \frac{2x}{5} +7>6\)  and \(\displaystyle -6> \frac{2x}{5} +7\)

\(\displaystyle \frac{2x}{5} >-1\)               \(\displaystyle -13> \frac{2x}{5}\)

\(\displaystyle x>-\frac{5}{2}\)                  \(\displaystyle -\frac{65}{2}> x\)

So, our solution is \(\displaystyle -32.5>x \ or -2.5< x\)