Prove $\Bbb{Z}[\sqrt{d}] /(a + \sqrt{d}) \cong\Bbb{Z}/n\Bbb{Z}$, where $n=|a^2-d|$.

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Let $a,d$ be integers with $d$ square free. Prove that $\mathbb{Z}[\sqrt{d}]/(a + \sqrt{d}) \cong \mathbb{Z}/n\mathbb{Z}$ where $n= |a^2- d|.$

I've tried attempting the problem by looking for a surjective homomorphism with kernel $(a + \sqrt{d})$. I thought about using the standard norm $N(x + y\sqrt{d}) = |x^2 - dy^2|$ but its not additive, I'm also not sure it's surjective.

This isn't homework, I'm just looking up practice problems to study for an exam. Any suggestions are greatly appreciated.