How many **real solutions** does the equation $x^{2} + 4x + 4 = 0$ have, and what are they?

The discriminant determines how many real solutions a quadratic has: $\Delta = b^{2} - 4ac = 4^{2} - 4(1)(4) = 16 - 16 = 0$. When $\Delta = 0$, there is exactly **one** real solution (a repeated root): $x = \dfrac{-b}{2a} = \dfrac{-4}{2} = -2$. Confirm by factoring: $x^{2} + 4x + 4 = (x + 2)^{2}$. Setting $(x + 2)^{2} = 0$ gives $x = -2$ as the only solution. Three-case rule for the discriminant: $\Delta > 0$ → two distinct real roots; $\Delta = 0$ → one repeated root; $\Delta < 0$ → no real roots.