Treat the '$\displaystyle x$' as a constant, and then substitute into the quadratic formula, or complete the square.
$\displaystyle 4y^2 - 5xy - 2x^2 + 1 = 0$
Compare this with the general form of a quadratic:
If $\displaystyle ay^2+by+c=0$
then $\displaystyle y=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here, your $\displaystyle a=4$, $\displaystyle b=-5x$ and $\displaystyle c=-2x^2+1$, so do the substitution.