The problem provides the information that AB,BC are the legs of right triangle and AC represents the hypotenuse, hence, you need to use Pythagora's theorem, such that:

`AC^2 = AB^2 + BC^2`

Replacing 48 for AC, `4x - 15` for AB and `2x + 3` for BC yields:

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

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

`(48 - (2x + 3))(48 + 2x + 3) = 16x^2 - 120x + 225`

`(45 - 2x)(51 + 2x) = 16x^2 - 120x + 225`

`2295 + 90x - 102x - 4x^2 = 16x^2 - 120x + 225`

`20x^2 - 108x - 2070 = 0 => 10x^2 - 54x - 1035 = 0`

You ened to complete the square `10x^2 - 54x` such that:

`a^2 = 10x^2 => a = xsqrt10`

`2ab = 54x => 2*x*sqrt10*b = 54 x`

`b = 27/sqrt10`

`10x^2 - 54x + 729/10 = 1035 + 729/10`

`(x*sqrt10 - 27/sqrt10)^2 = 11079/10`

`x*sqrt10 - 27/sqrt10 = +-sqrt(11079/10)`

`x*sqrt10 = (27+-sqrt11079)/sqrt10 => x = (27+-sqrt11079)/10`

You need to exclude `x = (27-sqrt11079)/10` as it creates a negative length of a leg.

**Hence, evaluating x, under the given conditions, yields **`x = (27+sqrt11079)/10.`

Since the given sides represent the cathetus and the hypothenuse of a right angle triangle, we'll determine x applying Pythagorean theorem:

hypothenuse^2 = cathetus^2 + cathetus^2

From enunciation, the hypothenuse is AC:

AC^2 = AB^2 + BC^2

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

We'll expand the square form the right side:

2304 = 16x^2 - 120x + 225 + 4x^2 + 12x + 9

We'll combine like terms and we'll use symmetric property:

20x^2 - 108x - 2070 = 0

10x^2 - 54x - 1035 = 0

We'll apply the quadratic formula:

x1 = [54+sqrt(44316 )]/20

x1 = (54+210.51)/20

x1 = 13.2255 approx.

x2 = -7.8255 approx.