We have:

\(2x^2+3y^2+4x=19\)

\(\Leftrightarrow2x^2+4x+2=21-3y^2\)

\(\Leftrightarrow2\left(x+1\right)^2=3\left(7-y^2\right)\)

Because \(2\left(x+1\right)^2⋮2\Rightarrow3\left(7-y^2\right)⋮2\)

But \(3⋮̸2\)=> \(7-y^2⋮2\)

=> y² is a odd => y is a odd

We have:

\(\left(x+1\right)^2\ge0\Rightarrow7-y^2\ge0\Rightarrow0\le y^2\le7\)

And y is a integer => \(y^2\in\left\{0;1;4\right\}\)

But y² is a odd => y² = 1 <=> \(y=\pm1\)

\(\Rightarrow2\left(x+1\right)^2=21-3\cdot1=18\)

\(\Rightarrow\left(x+1\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

So we have \(\left(x,y\right)=\left(2;1\right);\left(2;-1\right);\left(-4;1\right);\left(-4;-1\right)\)

We have:

2x2+3y2+4x=19

⇔2x2+4x+2=21−3y2

⇔2(x+1)2=3(7−y2)

Because 2(x+1)2⋮2⇒3(7−y2)⋮2

But 3⋮̸2

=> 7−y2⋮2

=> y2 is a odd => y is a odd

We have:

(x+1)2≥0⇒7−y2≥0⇒0≤y2≤7

And y is a integer => y2∈{0;1;4}

But y2 is a odd => y2 = 1 <=> y=±1

⇒2(x+1)2=21−3⋅1=18

⇒(x+1)2=9

⇒[x+1=3x+1=−3⇒[x=2x=−4

So we have (x,y)=(2;1);(2;−1);(−4;1);(−4;−1)