Solving general general cubic and quartic polynomials by radicals is a problem with an interesting history.
Anyways, if you want to solve your numerically, you can use Newton's method to get a real root:
x ≈ 2.72741
Factoring x - 2.72741... out of your cubic x3 + x2 - x - 25 = 0 gives a quadratic, which when set to 0 and solved by quadratic formula gives the other two roots, which are complex:
x ≈ -1.86371 ± 2.38596 i
If yo want to solve your cubic by radicals, you can do it as follows, since it fails the rational root test.
In the cubic ax3+bx2+cx+d=0, if you substitute x = y - b/(3a), expand and simplify by collecting terms with equal powers of y, you get a cubic polynomial in y in which there is no y2 term. We also need to divide both sides by a and put the linear and constant term on the right hand side. The resulting cubic is called a depressed cubic (general form is y3 = py + q).
With your cubic, x3 + x2 - x - 25 = 0, we let x = y - 1/(3*1) = y - 1/3 and substitute.
(y - 1/3)3 + (y - 1/3)2 - (y - 1/3) - 25 = 0
y3 - y2 + (1/3)y - 1/27 + y2 - (2/3)y + 1/9 - y + 1/3 - 25 = 0
y3 - (4/3)y - 664/27 = 0
y3 = (4/3)y + 664/27 (This is our depressed cubic equation).
Step 2: Solving the depressed cubic.
Observe the identity (s+t)3 = 3st(s+t) + s3+t3. I will leave it to you as an exercise to verify this identity.
Notice that it is of the form y3 = py + q where p = 3st, q = s3 + t3, and y =s+t.
Thus if we find two numbers p = 3st, q = s3 + t3, we will find that y = s + t is a root.
If we factor out y-(s+t) and solve the resulting quadratic by the quadratic formula, it turns out that the other two roots are
y = -(s+t)/2 ± (√3)(s-t)i/2.
With your cubic, for example, we derived the depressed cubic and saw that p = 4/3 and q = 664/27 so we seek numbers s and t satisfying 3st = 4/3 and s3 + t3 = 664/27.
In the first equation, dividing by 3 and cubing gives s3t3 = 64/729. Thus s3 and t3 are roots of a quadratic
u2 - (664/27)u +64/729 = 0
which when solved using the quadratic formula gives u = (332 ± 36√85)/27.
Taking cube roots give the expressions for s and t so we can back substitute in the formulas for y and then back substitute the y values to get x. Here is the final result:
x = (-1 + 3√(332 + 36√85) + 3√(332 - 36√85) )/3
x = -1/3 - (3√(332 + 36√85) + 3√(332 - 36√85))/6 + i(√3)(3√(332 + 36√85) - 3√(332 - 36√85))/6
x = -1/3 - (3√(332 + 36√85) + 3√(332 - 36√85))/6 - i(√3)(3√(332 + 36√85) - 3√(332 - 36√85))/6