How many **real solutions** does the equation $x^{2} + x + 5 = 0$ have?

Compute the discriminant: $b^{2} - 4ac = 1^{2} - 4(1)(5) = 1 - 20 = -19$. Because the discriminant is **negative**, the quadratic formula's $\sqrt{b^{2} - 4ac}$ is the square root of a negative number — which is not a real number. So the equation has **no real solutions**. Geometric reading: the parabola $y = x^{2} + x + 5$ never touches the $x$-axis. The vertex is at $x = -1/2$, where $y = (1/4) - (1/2) + 5 = 4.75 > 0$, so the entire parabola sits above the $x$-axis. The equation does have complex solutions ($x = \frac{-1 \pm i\sqrt{19}}{2}$), but the GRE asks for *real* solutions.