Trial and Improvement
Any equation can be solved by trial and improvement (/error). However, this is a tedious procedure.

Example:
Solve t³ + t = 17 by trial and improvement.

Firstly, select a value of t to try in the equation. I have selected t = 2. Put this value into the equation. We are trying to get the answer of 17.
If t = 2, t³ + t = 2³ + 2 = 10 . This is lower than 17, so we try a higher value for t.
If t = 2.5, t³ + t = 18.125 (too high)
If t = 2.4, t³ + t = 16.224 (too low)
If t = 2.45, t³ + t = 17.156 (too high)
If t = 2.44, t³ + t = 16.966 (too low)
If t = 2.445, t³ + t = 17.061 (too high)

So we know that t is between 2.44 and 2.445. So to 2 decimal places, t = 2.44.

Iteration
This is a way of solving equations.
An iteration formula might look like the following:
x_n+1  = 2  +  1
               x_n

You are usually given a starting value, which is called x₀. If x₀ = 3, substitute 3 into the original equation where it says x_n. This will give you x₁. (This is because if n = 0, x₁ = 2 + 1/x₀ and x₀ = 3).
x₁ = 2 + 1/3 = 2.333 333 (by substituting in 3).
To find x₂, substitute the value you found for x₁.
x₂ = 2 + 1/(2.333 333) = 2.428 571

Repeat this until you get an answer to a suitable degree of accuracy. This may be about the 5th value for an answer correct to 3s.f. In this example, x₅ = 2.414...

Example:
a) Show that x = 1 +   11    
                               x - 3
is a rearrangement of the equation x² - 4x - 8 = 0.

b) Use the iterative formula X_n+1 = 1 +   11  
                                                        x_n - 3
together with a starting value of x₁ = -2 to obtain a root of the equation x² - 4x - 8 = 0 accurate to one decimal place.

a) multiply everything by (x - 3):
x(x - 3) = 1(x - 3) + 11
so x² - 3x = x + 8
so x² - 4x - 8 = 0

b) x₁ = -2
x₂ = 1 +   11    (substitute -2 into the iteration formula)
            -2 - 3
    = -1.2
x₃ = 1 +      11        (substitute -1.2 into the above formula)
            -1.2 - 3
    = -1.619
x₄ = -1.381
x₅ = -1.511
x₆ = -1.439
x₇ = -1.478
therefore, to one decimal place, x = 1.5 .