Solve $y^2 + 3xy - 10x^2 + y + 5x = 0$ for y in terms of x

Question

I'm given the following equation: $y^2 + 3xy - 10x^2 + y + 5x = 0$ and asked to solve $y$ in terms of $y$.

My attempt:

$y^2 + (3x+1)\times y - 10x^2 + 5x = 0$

$\Rightarrow (y+(3x+1)/2)^2 - ((3x+1)/2)^2 = (x - (\frac 12)\times x)^2 - (\frac 12)^2 $

But I cant seem to get any further and it's probably in the wrong direction.

Answer 1

$$y^2 + 3xy - 10x^2 + y + 5x = 0\Rightarrow y^2+(3x+1)y-(10x^2-5x) = 0$$

So $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(3x+1)^2+4(10x^2-5x)}}{2}$$

$$\displaystyle = \frac{-(3x+1)\pm \sqrt{49x^2-14x+1}}{2}$$

so we get $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(7x-1)^2}}{3} = \frac{-(3x+1)\pm (7x-1)}{2}$$

So we get $\displaystyle y = 2x-1\;\;,-5x$

Answer 2

$0 = y^2+3xy-10x^2 + y+5x= (y+5x)(y-2x)+(y+5x) = (y+5x)(y-2x+1)$

Now this is much simpler to solve