For all integers b, c, and d, if x is rational such that x^2+bx+c=d, then x is an integer
This is simply the monic quadratic case of the Rational Root Test. You could specialize that proof, or else proceed similarly to various irrationality proofs for square-roots. $\: $ E.g. $\:$ below is a proof that I discovered in high-school. First I present the proof for square-roots - where the idea is clearer.
Theorem $\ $ For $\rm\: c\in \mathbb Z,\:$ any rational root $\rm\:r\:$ of $\rm\ x^2 = c\ $ is am integer.
Proof $\ $ Put $\rm\ \color{#0a0}{r = m/n}\ $ with $\rm\:(m,n) = 1.\:$ Then $\rm\ \color{#c00}{jm-kn =1}\;$ for some $\:\rm j,k \in \mathbb{Z}\,$ by Bezout.
Hence $\,\rm \color{#0a0}{0 = (m-nr)}\:(k+jr) = mk-njc + (\color{#c00}{jm-kn}) r \ \Rightarrow\ r = -mk+njc \ \in\ \mathbb{Z}\ \ \ $ QED
This proof easily extends to the root of a general monic quadratic as follows.
Theorem $\ $ For $\rm\:b,c\in\mathbb Z,\,$ any rational root $\rm\:r\:$ of $\rm\ x^2 = \color{#90f}{b\ x + c}\ $ is an integer.
Proof $\ $ Put $\rm\ \color{#0a0}{r = m/n},\ (m,n)=1,\,$ so $\rm\,(m\!-\!nb,n)=1\ $ so $\rm\, \exists\ j,k\in \mathbb Z\!:\ \color{#c00}{1 = j(m\!-\!nb)\!-\!kn} $
Hence $\rm\, \color{#0a0}{0 = (m\!-\!nr)}(k\!+\!jr)\ =mk\! +\! (jm\!-\!kn)r\!-nj(\color{#90f}{br\!+\!c}) = mk\!-\!njc + (\color{#c00}{j(m\!-\!nb)\!-\!kn})r$
The same proof easily extends to higher degree polynomials that are monic (lead coef $=1).$
If you learn about denominator ideals then you'll see that the above proof simply says that the denominator ideal of $\rm\:r\:$ contains $\rm\:n\:$ and $\rm\:nr = m,\:$ so it contains their gcd $\rm\:(n,m) = 1,\,$ so $\rm\ r\in \mathbb Z.$ Using Dedekind's notion of conductor ideal, the proof easily generalizes to higher degree monic polynomials, yielding that PIDs are integrally closed.
Hint: First, you can combine $c$ and $d$ as you only care about their difference. There is the theorem that the square root of a positive integer is either integer or irrational. You are right that one example disproves the assertion "for all b,c, and d".
Added: If you look at the quadratic formula, the only threat to x being integral (if it is rational) is the division by 2. But if the square root is integral, it must have the same parity as b.