There are many solutions to this. Here is one. The given sides are marked and labelled (1st side AB=77, 2nd side BC=107, 3rd side CE=149, last side AD=156). In this case x=DE=118.3 approx. The arbitrary positions of C, D and E determine x. In the absence of further information there is no unique value for x.
The interior angles of a pentagon sum to 540°, so if the given information had been angles rather than sides there is a unique solution, x=540-(77+107+149+156)=51°. (The central angles of a regular pentagon, formed by aplitting it into 5 isosceles triangles, are 72°. The other 2 angles sum to 180-72=108°. Two of these form each interior angle so the sum of all the interior angles is 5×108=540°. Irregular pentagons have the same sum for their interior angles.)