Your calculator's cubic regression function can tell you the equation is
... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)
The x-intercepts are -4, 0, +1.5.
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If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as
... f(x) = ax³ + bx² + cx
Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.
... -a +b -c = 15 . . . . . for x=-1
... a + b + c = -5 . . . . for x=1
... 8a +4b +2c = 12 . . for x=2
adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to
... a + c = -10
... 4a +c = -4
Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).
Then your equation is
... f(x) = x(2x² +5x -12)
Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.
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Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have
... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above
or 
or x=
