Use Heron's formula.

For any triangle XYZ, its area is equal to

√(s(s-x)(s-y)(s-z)), where s =

For triangle ABC, s = 3x + 3

Area = √((3x+3)(2x-1)(3-x)(2x+1))

For triangle ADC, s = 2x + 13

Area = √((2x+13)(2x+1)(3-x)(x+9)}

So, we want to solve the inequality

√((3x+3)(2x-1)(3-x)(2x+1)) > √((2x+13)(2x+1)(3-x)(x+9)}

It is compulsory that each of the multiplied terms in parentheses be greater than zero. If - a were less than or equal to zero, we can derive a >= b + c, which violates the triangle inequality. We therefore can cancel out like terms while preserving the inequality - that is,

√((3x+3)(2x-1)) > √((2x+13)(x+9))

From triangle ABC, we are given

2x-1 > 0 -> x > 0.5 and
3-x > 0 -> x < 3

Quite conveniently, this guarantees that (2x+13)(x+9) > 1, so the special cases of √n > √m while n and m can be similtaneously between zero and one is irrelevant, and we can proceed directly to

2x² + 31x + 117 < 6x² + 3x - 3 ->

4x² - 28x - 120 > 0 ->

x² - 7x - 30 > 0

This is factorable.

(x-10)(x+3) > 0 ->

x>10 or x<-3

The intersection of this solution and the compulsory 0.5
Answer: Ø