User:Satori/proof

$$\frac{\sqrt{2}-\sqrt{6}}{4} = \frac{-\sqrt{2-\sqrt{3}}}{2}$$

Multiplying by two:

$$\frac{\sqrt{2}-\sqrt{6}}{2} = -\sqrt{2-\sqrt{3}}$$

Squaring both sides:

$$\frac{2-2(\sqrt{2}\sqrt{6})+6}{2^2} = 2-\sqrt{3}$$

Simplified:

$$\frac{8-2\sqrt{12}}{4} = 2-\sqrt{3}$$

Multiplying by four:

$$8-2\sqrt{12} = 8-4\sqrt{3}$$

Subtracting eight and multiplying by negative one:

$$2\sqrt{12}= 4\sqrt{3}$$

Factoring a two from the right side and representing as the square root of four:

$$2\sqrt{12} = 2\sqrt{4}\sqrt{3}$$

$$2\sqrt{12} = 2\sqrt{12}$$