AB = 4x (AC = hypotenuse )

AB =4x-15 and BC = 2x+3. AC = 48. AC is the
hypotenuse.

To find the value of
x.

From the given details, ABC is a right angled triangle with AC as
hypotenuse . Therefore AB and AC form a right angle at B.

So
AB^2+BC^2 = AC^2  ....(1) by  Pythagoas theorem.

We substitute the
given values AB = 4x - 15 and BC = 2x + 3 and  AC = 48 in (1) and solve the quadratic equation
for x. :

(4x-15)^2 +(2x+3)^2 =
48^2.

16x^2 -120x +225 +4x^2+12x +9 - 48^2 =
0

20x^2-108x - 2070 = 0....(1)

We use
the quadratic formula : The solution of ax^2+bx+c= 0 is x1 = {-b +or- sqrt(b^2-4ac)}/2a. Here a=
20, b = -108 and c = 2070.

So x1 = {108
+sqrt(108^2-4*20(-2070)}/2*40 = 13.2257 nearly.

x2 =  {108
+sqrt(108^2-4*20(-2070)}/2*40 = -7.8257 nearly.