Sometimes I grind through the quadratic formula only to end up with a negative under the root.
Accepted answer
Compute the **discriminant** — the part under the radical in the quadratic formula:
**D = b² − 4ac**
- `D > 0` → **two** distinct real solutions (parabola crosses the x-axis twice)
- `D = 0` → **one** repeated real solution (vertex sits on the x-axis)
- `D < 0` → **no** real solutions; two complex conjugates (parabola misses the x-axis)
For `x² + 2x + 5`: `D = 4 − 20 = −16 < 0`, so no real solutions — and you knew before doing any work.
The sign is what matters, so you can often stop early. It's also how questions like "for what values of `k` does this have exactly one solution?" are meant to be answered: set `D = 0` and solve for `k`.