A correct solution was submitted by Kirstin Jerzy.  Further correct solutions were sent in by Philippe Fondanaiche, France; Bill Webb, USA; Farid Lian, Colombia; Ron Welch, USA; John Snyder, USA; Juan Carlos Marivela, Spain.

The  quadratic equation has a real root if and only if its discriminant (the "b² - 4ac" part of the quadratic formula) is non-negative. Writing D for this discriminant, we have 

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which simplifies to a + 2d ≥ (a + d)²,  a ≠ 0, a ≠ -d, a ≠ -2d.  

What planar region does this describe?  Write this as (a + d) + d ≥ (a + d)² and set  x = a + d, y = d; the equation becomes x + y ≥ x², or  y ≥ x² - x which describes the "inner" part of the parabola having roots 0 and 1.  The original equation is then a parabola, rotated by 45°; see the figure on the left.  Returning to the original equation, and solving for b in terms of a gives b = (1 - a) ± √(1 - a).  Our inequality has a real solution if and only if the pair (a,b) lies in the shaded region.  The restriction that a and d be positive gives the solution to the original quadratic:

0 ≤ a ≤ 1  and  0 ≤ b ≤  (1 - a) + √(1 - a).

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