Wait... you want me to solve this in whole numbers?

Integers

If $x=0$ then $2y^2+7y+6=0.$ The only acceptable solution in integers is $(x,y)=(0,-2).$ If $y=0$ then there is no integer solution in $x.$ Now let $xy\ne 0.$ Treat the given expression as a quadratic in $x$. Its discriminant is $\Delta=5(85y^4+160y^3-94y^2+4y-43).$ Therefore we must have $94y^2-4y+43=0\mod 5$ or $y^2-y+2=0\mod 5.$ But that's not possible, so the only solution is $(x,y)=(0,-2).$

tk1 wrote: If $x=0$ then $2y^2+7y+6=0.$ The only acceptable solution in integers is $(x,y)=(0,-2).$ If $y=0$ then there is no integer solution in $x.$ Now let $xy\ne 0.$ Treat the given expression as a quadratic in $x$. Its discriminant is $\Delta=5(85y^4+160y^3-94y^2+4y-43).$ Therefore we must have $94y^2-4y+43=0\mod 5$ or $y^2-y+2=0\mod 5.$ But that's not possible, so the only solution is $(x,y)=(0,-2).$ Discriminant is $\Delta=425y^4+720y^3+270y^2+60y+25$ not $\Delta=5(85y^4+160y^3-94y^2+4y-43).$

We have given the Diophantine equation $(1) \;\; 5x^2y^2 + 10x^2y + 25xy^2 + 30xy + 5x + 2y^2 + 7y + 6 = 0$. Equation (1) is equivalent to $(2) \;\; (25x^2 + 25x + 2)y^2 + (10x^2 + 30x + 7)y + 5x + 6 = 0$. Let us consider the three cases $x \in \{-2,-1,0\}$: $\bullet \: x=-2 \;\;\; \Rightarrow \;\;\; 52y^2 - 13y + 4 = 0 \;\;\; \Rightarrow \;\;\; 13|4$ (contradiction) $\bullet \: x=-1 \;\;\; \Rightarrow \;\;\; 2y^2 - 13y + 1 = 0 \;\;\; \Rightarrow \;\;\; |y|=1$ (impossible) $\bullet \: x=0 \;\;\; \Rightarrow \;\;\; 2y^2 - 7y + 6 = 0 \;\;\; \Rightarrow \;\;\; (2y + 3)(y + 2) = 0 \;\;\; \Rightarrow \;\;\; y=-2$. Assume $x \leq -3$ or $x \geq 1$. Let $f(x,y)$ by the LHS of equation (2). If $y>0$, then $0 = f(x,y) \geq f(x,1) = 35x^2 + 60x + 8 > 0$ (since $x \neq -1$), a contradiction implying $y \leq 0$. If $y=0$, then $x=-1.2 \not \in \mathbb{Z}$ by equation (2). Hence $y \leq -1$, yielding $y^2 \geq -y$, which combined with the facts $25x^2 + 25x + 2 > 0$ and $10x^2 + 30x + 7 > 0$ (since $x \neq -2$ and $x \neq -1$) give us $0 = f(x,y) \geq (25x^2 + 25x + 2)(-y) - (10x^2 + 30x + 7)(-y) + 5x + 6 = (15x^2 - 5x - 5)(-y) + 5x + 6 \geq 15x^2 + 1 > 0$, a contradiction which means equation (1) has no solution when $y \leq 0$. Conclusion:. The only solution of equation (1) is $(x,y) = (0,-2)$.

The problem formulation written by now is wrong! Here is the correct one https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2018/06/20171.pdf and here goes a solution sketch. As a quadratic with respect to $y$, we have \[ (25x^2 + 25x + 2)y^2 + (10x^2 + 30x + 7)y + x^2 + 5x + 6 = 0. \]The discriminant is $D = (10x^2 + 30x + 7)^2 - 4(x^2+5x+6)(25x^2+25x+2) = -68x^2 - 220x + 1 = 1 - 4x(17x + 55)$. It is negative for $x\geq 1$ and $x\leq -4$ and for $x=-3,-2,-1,0$ just solve the equation separately. The solutions are $(-3,0)$, $(-2,0)$, $(0,-2)$.