An algebra problem by Joe Mansley

Solving for x via the Quadratic Formula yields:

$x = \frac{(10y + 10) \pm \sqrt{(10y+10)^2 - 4(5)(10y^2 - 10y + 25)}}{2(5)} = \frac{(10y+10) \pm \sqrt{-100y^2 + 400y - 400}}{10} = \frac{(10y+10) \pm \sqrt{-100(y-2)^2}}{10} = \frac{(10y+10) \pm 10i(y-2) }{10}$ ;

or $x = (y+1) \pm (y-2)i$ .

If $x \in \mathbb{R}$ , then this forces $y = 2 \Rightarrow \boxed{x=3}.$