Number of solutions to a polynomial I need to find the number of solution to the polynomial 

$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)=0$$

Clearly, $x=1$ and $y=0$ is a solution. How to prove it has no other solutions or more solutions?

Considering the previous answers, I could be totally wrong.
If the equation is 
$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y\color{red}{+}4)=0$$
expand it  as a polynomial of $x$ to get
$$21 x^2- (39 y+34)x+(19 y^2+23 y+13)=0$$ Solve the quadratic to get
$$x_\pm=\frac{1}{42} \left(39 y+34\pm\sqrt{-75 y^2+720 y+64}\right)$$ and notice that 
$$-75 y^2+720 y+64 \geq 0 \qquad \text{if}\qquad \frac{8}{15} \left(9-2 \sqrt{21}\right)\leq y \leq \frac{8}{15} \left(9+2 \sqrt{21}\right)$$ which seems to mean that there is an infinite number of solutions in the given range for $y$ which seems normal since the equation corresponds to a rotated ellipse.
