How do we know that these roots are non-real? The roots, x and y, of a quadratic equation $$ax^2 + bx + c=0$$ satisfy the following inequality:
$$x^2 + y^2 < 0$$
What can you conclude about the nature of the roots of the quadratic equation?

$$x^2 + y^2 = (x+y)^2 - 2xy = \frac{b^2 - 2ac}{a^2}$$
since $$x^2 + y^2 <0$$, we can say $$b^2 - 2ac < a^2$$
therefore, $$b^2 - 4ac < a(a-2c)$$
therefore,$$ D < a(a-2c)$$
How do I proceed from here?

Alternatively:
$$x^2+y^2=\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2=\frac{b^2-4ac}{a^2}<0 \Rightarrow$$
$$b^2-4ac<0 \Rightarrow x,y\in \mathbb{C}.$$
