解：对于方程$x^2 - 3x - 1 = 0，$其中$a = 1,b = -3,c = -1，$根据求根公式$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}，$可得$x=\frac{3\pm\sqrt{(-3)^2 - 4\times1\times(-1)}}{2\times1}=\frac{3\pm\sqrt{13}}{2}，$所以$x_1=\frac{3 + \sqrt{13}}{2},x_2=\frac{3 - \sqrt{13}}{2}$
解：对于方程$x^2 + 4x - 4 = 0，$其中$a = 1,b = 4,c = -4，$根据求根公式$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}，$可得$x=\frac{-4\pm\sqrt{4^2 - 4\times1\times(-4)}}{2\times1}=\frac{-4\pm\sqrt{32}}{2}=-2\pm2\sqrt{2}，$所以$x_1 = 2\sqrt{2}-2,x_2=-2\sqrt{2}-2$
解：对于方程$2x^2 - 2\sqrt{2}x=-1，$变形为$2x^2 - 2\sqrt{2}x + 1 = 0，$其中$a = 2,b=-2\sqrt{2},c = 1，$根据求根公式$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}，$可得$x=\frac{2\sqrt{2}\pm\sqrt{(-2\sqrt{2})^2 - 4\times2\times1}}{2\times2}=\frac{2\sqrt{2}\pm0}{4}=\frac{\sqrt{2}}{2}，$所以$x_1 = x_2=\frac{\sqrt{2}}{2}$
解：对于方程$4x(x - 1)=6 - 6x，$变形为$4x(x - 1)+6(x - 1)=0，$提取公因式得$(x - 1)(4x + 6)=0，$则$x - 1 = 0$或$4x + 6 = 0，$解得$x_1 = 1,x_2 = -\frac{3}{2}$
解：①设$x^2 = y，$则原方程可化为$y^2 - y - 6 = 0，$因式分解得$(y - 3)(y + 2)=0，$解得$y_1 = -2,y_2 = 3.$ 当$y = -2$时，$x^2 = -2，$无实数根；当$y = 3$时，$x^2 = 3，$解得$x=\pm\sqrt{3}.$ 所以原方程的根为$x_1=\sqrt{3},x_2 = -\sqrt{3}$
解：②设$x^2 + 3 = a，$则原方程可化为$a^2 - 9a + 20 = 0，$因式分解得$(a - 4)(a - 5)=0，$解得$a_1 = 4,a_2 = 5.$ 当$a = 4$时，$x^2 + 3 = 4，$解得$x=\pm1；$当$a = 5$时，$x^2 + 3 = 5，$解得$x=\pm\sqrt{2}.$ 所以原方程的根为$x_1 = 1,x_2 = -1,x_3=\sqrt{2},x_4 = -\sqrt{2}$
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