• d > 0 yields 2 distinct roots
• d = 0 gives repeated roots
• d < 0 means there aren't any real number solutions. The two solutions are complex.

    (1)  -x^2 + 18x + 81
    (2)  3x^2 - 6x + 9
    (3)  8x^2 - 32x + 32
    (4)  25x^2 - 30x - 9
    (5)  x^2 - 14x + 196

If a quadratic polynomial has repeated root, then the polynomial is

    +/- (ax-b)^2,  or  +/- (a^2 - 2abx + b^2).


We see that if a quadratic polynomial has repeated root, then the coefficient at x^2 and the constant term
have the same sign: they either both are positive or both are negative,


So, cases (1) and (4) fall out;  cases (2), (3) and (5) remain for the further consideration.


In case (5),  the coefficient at x should be  +/- 2*sqrt(coef at x^2)*sqrt(constant term) = +/- 2*sqrt(1)*sqrt(196) = +/- 28,
              but we have there number -14 instead, so case (5) falls out.  Cases (2) and (3) remain for further consideration.


In case (3) ,  8x^2 - 32x + 32 = 8(x^2 -4x + 4) = 8*(x-2)^2  has the repeated root 2,
               so, it is that unique case, we are searching for.


At this point, the unique case is just found; so, case (2) does not require further analysis.


ANSWER.  The unique polynomial with repeated root is case (3).

         The repeated root is 2.