The Formula Goes Complex

When ${b}^{2}-4ac$ is negative, the square root in the quadratic formula holds a negative number. In Algebra that meant no real solution; now it means two complex solutions. Take ${x}^{2}-6x+34 = 0$: a is 1, b is -6, c is 34, and ${b}^{2}-4ac$ is $36-136$, which is -100.

So $x = \frac{6\pm\sqrt{-100}}{2}$, and $\sqrt{-100}$ is 10i, giving $x = \frac{6\pm10i}{2}$. Divide both terms of the top by 2: $x = 3\pm5i$. Split the ±: $x = 3+5i$ or $x = 3-5i$.

When a, b and c are real, the two always come as a conjugate pair: the real part, b divided by 2a with its sign flipped, is shared, and the ± sits only in front of the i term.