Once the powers of x are arranged into the following form (an through a0 are known constants), the formula can be solved for values of x that satisfy the equation:
anxn + an-1xn-1 + an-2xn-2 + .... + a2x2 + a1x + a0
For example, if n = 4, it will be a fourth order equation, so there
will be four values for x.
Input values for all the 'a' terms below, and then click the "Calculate Solutions" button to get the results. (The form has been pre-filled as a fourth order polynomial. This is a special case where all four roots are the same, ie. -1. Examples of special cases like this are when the coefficients are numbers from a Pascal Triangle.)
NB: As higher order polynomials are solved, small inaccuracies will
creep in - pushing it beyond 6th order will make these progressively more
significant. They should be interpolated accordingly.
The solutions might include imaginary components,
as indicated by the "j" field.
Note: if a0 is zero, one of the roots found
will be zero because the equation is really one order less.
Likewise, if a1 is also zero, two roots
will be zero because the equation is really two orders less, and so on.