After 16 literations it gives the answer of 2.627 to 3 d.p or 2.627 +/ 0.0005. Therefore it can be said that that 2.627<X<2.628.
As the sign of (X) when its equal to 2.627 is negative
i.e. 2.627 3 – 5*2.627  5 = 0.00573
 And the sign of (X) when it equals 2.628 is positive
i.e. 2.628 3 – 5*2.628  5 = 0.00998
To give an example of an equation that cannot be solved using this method is an equation that follows two main points:

It must be an equation that just touches the xaxis but doesn’t cross the axis i.e. X4 (shown on omnigraph below).

Must not touch the xaxis at the origin therefore the graph of X4 needs to be ‘shifted’ either left or right i.e. (X7)4 [shown below on omnigraph].
This highlights how there is no solution using this method as it relies on the equation crossing the xaxis not just touching as this equation does (X7)4.
The Newton Raphson Method.
This shows two solutions to the equation the first I intend to find lies between 0 and 1 therefore an estimate of the root for 2X4 X3 + 3X22 = 0 is given by (0.5). A tangent is then drawn to the curve y = 2X4 X3 + 3X22. The point where the tangent crosses the xaxis is the next estimate for the root and the process repeated. Below shows the soluiton to the first root using Auto Graph.
By using excel I setup a spread sheet to prove Auto graph gave the correct solution at 0.7661. I then used this method to calculate the other root estimated at 0.5
This shows that there are two solutions to the equation: 2X4 X3 + 3X22 = 0
It shows that after 4 iterations it gives one answer of 0.766097 to 6 d.p or 0.766097 +/ 0.0000005. Therefore it can be said that that 0.766096<X<0.766097.
As the sign of (X) when its equal to 0.766096 is negative
i.e. 2*0.7660964 0.7660963 + 3*0.76609622 = 4.4 x 1006
 And the sign of (X) when it equals 0.766097 is positive
i.e. 2*0.7660974 0.7660973 + 3*0.76609722 = 2.03 x 1006
Also after 4 iterations it gives another solution of 0.66344 to 5 d.p or 0.66344 +/ 0.000005. Therefore it can be said that that 0.66344<X<0.66345
A Nonsolve able equation using Newton Raphson.
An equation that’s solution can’t be solve/found is an equation where the point crosses the xaxis and is also very close to a maximum or minimum point. Therefore the tangent can’t be drawn to the curve as it is interfered with by the turning point on the graph and therefore instead of converging towards a root it diverges from the solution instead of tending towards.
An example of this is: X33X2+X+1 = 0.
The diagram below shows that the gradient of the curve X33X2+X+1 = 0 although is not too steep to allow the Newton Raphson method to calculate the value of the root but highlights how the solution lies close to a maximum point on the graph.
It is also clear that by using excel to create a spread sheet, as used to calculate the solution for: 2X4 X3 + 3X22 = 0, shows how the tangent and therefore the answer moves further away and doesn’t give a correct solution.