You need to open the brackets such that:

`x^4 + x^3 - 6x^2 - 3x^3 - 3x^2 + 18x + 4x^2 + 4x - 24 = 24`

You need to group like terms such that:

`x^4 - 2x^3 - 5x^2 + 22x - 48 = 0`

You need to notice that x = 3 is a root for the quartic polynomial such that:

`x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(ax^3 + bx^2 + cx + d)`

Opening the brackets yields:

`x^4 - 2x^3 - 5x^2 + 22x - 48 = ax^4 + bx^3 + cx^2 + dx - 3ax^3 - 3bx^2 - 3cx - 3d`

Equating the coefficients of like powers yields:

`a=1`

`b-3a=-2 => b = 1`

`-3d = -48 => d = 16`

`d-3c=22 => -3c = 6 => c = -2`

`x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(x^3 + x^2 - 2x + 16)`

You need to solve for x the equation `x^3 + x^2 - 2x + 16 = 0 ` such that:

`x^3 + x^2 - 2x + 16 = 0`

You should come up with the following substitution 3`y = 3x + ` 1 such that:

`(y-1/3)^3 + (y-1/3)^2 - 2(y-1/3) + 16 = 0`

Expanding the cube and the square yields:

y^3 - y^2 + y/9 - 1/27 + y^2 - 2y/3 + 1/9 - 2y + 2/3 + 16 = 0

Combining like terms yields:

`y^3 - 7y/3 + 452/27 = ` 0

`27y^3 - 63y + 452 = 0`

Notice that there are no real solutions to equation `27y^3 - 63y + 452 = 0` , hence, the given equation `(x^2 -3x+4)(x^2+x-6)=24 ` has two real solutions, `x = 3` and `x in (-3,-3.5)` and two complex conjugate solutions.