Solution for a quadratic equation The following is the solution from my book, but without the calculation path.
$$
\frac{(18 \pm \sqrt{-180})}{6} = 3 \pm\sqrt{-5} 
$$
Why is it $-5$?
 A: We have this:
$$\frac{18\pm\sqrt{-180}}6 = \frac{18}{6}\pm\frac{\sqrt{-180}}{6} = 3\pm\sqrt{\frac{-180}{36}}=3\pm\sqrt{-5}$$
A: $$\begin{align}
\frac{1}{6}(18\pm\sqrt{-180}) 
&= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-180}\\
&= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-4 \times 9 \times 5}\\
&=\frac{1}{6} \times (18) \pm \frac{1}{6} \times (2 \times 3) \times\sqrt{-5}\\
&= 3 \pm\sqrt{-5}
\end{align}$$
A: Note that no ingenuity is required. For clearing denominators reduces it to proving the equivalent equation $\:\sqrt{-180} = 6\ \sqrt{-5}\:,\:$ whose proof should now be obvious. In the same way, one can often reduce less trivial equations to a "simpler" form. This is especially helpful when one is attempting to follow proofs involving special functions whose laws may not be so familiar. Then, with luck, one may succeed in reducing an equation to a known basic law of the special function (e.g. an addition law). You may have already encountered such transformation-based proof strategies when attempting to prove trigonometric identities in high school. 
In fact this was a method that I employed in high school to help prove trig identities. On the left, starting from the identity to be proved, I worked forwards, applying known laws. On the right, starting from known trig identities, I worked backwards, hoping to connect the two chains somewhere in the middle - hence concluding the sought proof. When there are many possible transformations to apply at each step (i.e. a large branching factor) this method of simultaneously forward and backward chaining - combined with intelligent pruning - can help to constrain combinatorial explosion due to the exponential growth of the search tree.  My high school teacher liked this method so much that he incorporated it into his lessons. Later, when I arrived at MIT, I learned that such simultaneous forward and backward chaining is a standard technique for searching such trees. For example, it proves helpful in solving certain chess problems. It is always helpful to keep such "meta-level" techniques in mind when searching for proofs.
A: $\sqrt{-180} = \sqrt{36\cdot(-5)} = \sqrt{36}\cdot\sqrt{-5} = 6\sqrt{-5}.$
(I think Asaf Karagila's answer is somewhat more complicated than it needs to be.)
