`x^2-y^2+3x+4y-2=0`

`⇔4x^2-4y^2+12x+16y-8=0`

`⇔(4x^2+12x+9)-(4y^2-16y+16)-1=0`

`⇔(2x+3)^2 - (2y-4)^2=1`

`⇔(2x-2y+7)(2x+2y-1)=1`

Khi đó: `(2x-2y+7); (2x+2y-1) ∈Ư(1)={±1}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{2x-2y+7}&\text{1}&\text{-1}\\\hline \text{2x+2y-1}&\text{1}&\text{-1}\\\hline \text{x}&\text{-1}&\text{-2}\\\hline \text{y}&\text{2}&\text{2}\\\hline\end{array}

Vậy, nghiệm nguyên `(x;y)` của phương trình là:  `(-1;2)` ; `(-2;2)`

`#` `Uh`