can the sum of interior angles of a polygon be 4300 justify

No.
Let the number of sides of the polygon whose sum of interior angles is 4300 be n
now, sum of interior angles of a polygon is given by (n-2) * 180    //where n = number of sides//
so, (n-2) * 180 = 4300
or n-2 = 4300/180 = 23.88
or n = 23.88 + 2 = 25.88
but, number of sides cannot be a decimal number,
so, sum of interior angles of a polygon cannot be 4300