How to solve for $y$: $x^2-3xy+y^2-x+4=0$ Please, I want to know the method to arrive at the solution $y=\dfrac{-3x \pm \sqrt{5x^2+4x-16}}{2}$ , step by step.
Thank you in advance.

$$\text{Hint: }$$
$$
\\x^2 - 3xy + y^2 - x + 4 = 0 \\ \Updownarrow \\ y^2 + (-3x)y +(x^2 -x+4) = 0
$$
$$$$
$$\text{Now solve the quadratic equation with: }$$
$$\begin{align}
a &= 1 \\ b &= -3x \\ c &= (x^2 - x + 4)
\end{align}$$
