Answer:
{−1.47596288535, 1.73968983949, 9.73627304586}
Explanation:
The roots of this cubic are real and irrational, as indicated by a graph of it. (Rational roots would be divisors of 25.) The same graphing calculator can provide an iterative solution to 12 significant figures. Here, we have used Newton's method iteration. The iteration function is ...
g(x) = x -p(x)/p'(x) . . . where p'(x) is the derivative: 3x^2-20x
We started each iteration using the value shown on the graph.
x ∈ {−1.47596288535, 1.73968983949, 9.73627304586}
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The second attachment shows another calculator's solution. The values of x that are roots are the opposite of the constant in each binomial factor. You will notice this solution has a couple more significant figures than the one shown above.
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Additional comment
There are several formulas for finding the "exact" roots of the cubic. Here's a trig approach that works reasonably well for cubics with 2 or 3 real roots.
We can define, for p(x) = x³ +ax² +bx +c, ...
s = -a/3 = 10/3
t = b -a²/3 = -100/3
r = a(2a² -9b)/27 +c = -10(200)/27 +25 = -1325/27
d = √(-4t/3) = √(400/9) = 20/3
h = 4r/d³ = -53/80
Now, the roots are ...
x = s +d·sin(arcsin(h)/3 +2nπ/3) . . . . for n=-1, 0, 1
= 10/3 +20/3·sin(arcsin(-53/80)/3 + n(2π/3)) . . . . . "exact solution"
= (10/3)(1 +2·sin(-13.8303° +n(120°))
= -1.47596..., or 1.73969..., or 9.73627...