Integer solutions to $2x^2+5x+y^2=19$

Question

$$2x^2+5x+y^2=19$$

Don't know how to approach the problem. Similar equations required factoring after the completing a square or a similar trick. I don't see the possibility of that here though. Hints? Answers?

Answer 1

Note that $$\begin{align}19-2x^2-5x=y^2\ge 0&\Rightarrow 2x^2+5x-19\le 0\\&\Rightarrow \frac{-5-\sqrt{177}}{4}\le x\le\frac{-5+\sqrt{177}}{4}\\&\Rightarrow -4\le x\le 2\end{align}$$

Answer 2

The equation with small numbers allows easy calculations. As a simple quadratic equation we have $4x = -5 ±\sqrt{177-8y^2}$. There are just four possibilities: y = ±1, ±2, ±3, ±4 and only ±1 and ±4 are good. Thus $4x = -5 ±13$ and $4x = -5 ±7$. Finally the only solutions are (x, y) = (2, ±1) and (-3, ±4)