Prove that $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\frac{-b+\sqrt{D}}{2a})$

Question

Prove that $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\frac{-b+\sqrt{D}}{2a}) \quad a,b \in \mathbb{Z}$
My work:
Let $\alpha = \frac{-b+\sqrt{D}}{2a}$ $$\mathbb{Q}(\sqrt{D}) = \{x+y\sqrt{D}\mid x,y\in \mathbb{Q}\} $$ $$\mathbb{Q}(\alpha_1) = \{ x + y\alpha =x-\frac{yb}{2a} + \frac{y\sqrt{D}}{2a} \mid x,y \in \mathbb{Q} \}$$ We want to prove that two set are equal. Because of that we need to prove two inclusion.
Let $x-\frac{yb}{2a} + \frac{y\sqrt{D}}{2a}$ be element from $\mathbb{Q}(\alpha_1)$. Let also $t = x-\frac{yb}{2a}$ i $u = \frac{y}{2a}$. We know that elements $t$ i $u$ are elements of set $\mathbb{Q}$. So we conclude that element of set $\mathbb{Q}(\alpha_1)$ can be written as element of set $\mathbb{Q}(\sqrt{D})$.
How to show other inclusion?

Answer 1

Hint: $$x+y\sqrt{D}=x+by+2ay\left(\frac{-b+\sqrt{D}}{2a}\right)$$