$ x^2+\sqrt{d}x+d = 0 $ ,
Where $d$ is the real number whose integer part is $0$ and it's $n$-th decimal digit is given by (for $n>0$):
$ \text{NthDecimalDigit}(n) = \left \{ \begin{array}{l}0,\: \:\: \text{if $2n+2$ can be expressed as the sum of two primes}\\ 1, \:\: \: \text{otherwise.} \end{array} \right.$
Obviously, the discriminant is -3d and d by definition is non-negative. Therefore, it holds that:
The equation has a real solution (x=0) if and only if d=0 (1)
But obviously, it also holds that:
d = 0 if and only if Goldbach's conjecture is true.
It is commonly stated that "the quadratic formula can solve every quadratic equation" (I provide a proof that this is not the case). Anyway, what do you think is the strict answer to the question of the title (some scientists consider equivalence (1) as a "solution")?
Let $$x^2 + px + q = 0$$ a quadratic equation then the discrimant is given by $\sqrt{\frac{p^2}{4} - q}$ so in your case it's $\sqrt{\frac{d}{4} - d}$
Because $d>0$ it holds $\frac{d}{4} - d < 0$ so your equation has no solution.