Q.394. Quadrilateral in (3, 4, 5) triangle

Question 394.
Find the exact value of x as shown in this image.

Answer 394.
DE^2 = 7x^2 = DF^2 + EF^2
=> ∠ F is a rt. angle

Let ∠ AFD = F
=> ∠ CFE = 90° - F

Applying sine rule to Δs ADF and CEF,
sinA/x√2 = sinF/(3-x√3) and
cosA/x√5 = cosF/(4 - 2x)
=> sinF = (4/5) (3 - x√3)/x√2
and cosF = (3/5) (4 - 2x)/x√5

Squarring and adding,
(16/25) (3 - x√3)^2/(2x^2) + (9/25) (4 - 2x)^2/(5x^2) = 1
Using Wolfram Alpha,
Wolfram Alpha Link to answer
x = (6/31) [12 + 20√3 - √{10(91 + 48√3}]
≈ 0.950539. ... [The second value given is redundant as AC = 5].

I did some further Wolfram Alpha investigation to confirm the correctness of the above answer and found AF = 1.60866 and FC = 3.39134 to confirm that AF + FC = 5.
Wolfram Alpha Link

Link to YA!